How do I set a random number of random lowercase characters to a Struct member using memset in C
I am forced to use memset and drand48() to set a random number (2 - 7) of random characters that are lower case letters ('a' to 'z'). My code returns non ASCII characters and I am not sure why.
struct Record {
int seqnum;
float threat;
unsigned int addrs[2];
unsigned short int ports[2];
char dns_name[NUMLTRS];
};
My code is in a for loop:
memset(rec_ptr[i].dns_name, (char)((122 * drand48()) + 97),
((sizeof(char) * 7) * drand48()) + (sizeof(char) * 2));
c random memset
asked Nov 21 '18 at 19:19
jch3jch3
11
|
show 7 more comments
I am forced to use memset and drand48() to set a random number (2 - 7) of random characters that are lower case letters ('a' to 'z'). My code returns non ASCII characters and I am not sure why.
struct Record {
int seqnum;
float threat;
unsigned int addrs[2];
unsigned short int ports[2];
char dns_name[NUMLTRS];
};
My code is in a for loop:
memset(rec_ptr[i].dns_name, (char)((122 * drand48()) + 97),
((sizeof(char) * 7) * drand48()) + (sizeof(char) * 2));
c random memset
asked Nov 21 '18 at 19:19
jch3jch3
11
2
ASCII's range is only0-127. Unlessdrand48()returns0in(char)((122 * drand48()) + 97), it won't fit.
– Fiddling Bits
Nov 21 '18 at 19:20
1
Why are you forced to usememset? Doingdns_name[i] = ...seems more logical if you have to bang out multiple characters.
– tadman
Nov 21 '18 at 19:23
1
sizeof charis1per definition. Always. And1is neutral for multiplication, so drop it.lrand48()would be better fitting since you want integers.
– Swordfish
Nov 21 '18 at 19:24
2
. o O ( everytime one uses magic numbers, a kitty dies :( :( )
– Swordfish
Nov 21 '18 at 19:40
1
The problem withmemsetis that it's a very good way of setting multiple bytes to the same value, but a very poor way of setting many bytes to different values. In this case I think it would be far simpler to use one array assignment for each letter.
– Tim Randall
Nov 21 '18 at 19:59
|
show 7 more comments
I am forced to use memset and drand48() to set a random number (2 - 7) of random characters that are lower case letters ('a' to 'z'). My code returns non ASCII characters and I am not sure why.
struct Record {
int seqnum;
float threat;
unsigned int addrs[2];
unsigned short int ports[2];
char dns_name[NUMLTRS];
};
My code is in a for loop:
memset(rec_ptr[i].dns_name, (char)((122 * drand48()) + 97),
((sizeof(char) * 7) * drand48()) + (sizeof(char) * 2));
c random memset
asked Nov 21 '18 at 19:19
jch3jch3
11
I am forced to use memset and drand48() to set a random number (2 - 7) of random characters that are lower case letters ('a' to 'z'). My code returns non ASCII characters and I am not sure why.
struct Record {
int seqnum;
float threat;
unsigned int addrs[2];
unsigned short int ports[2];
char dns_name[NUMLTRS];
};
My code is in a for loop:
memset(rec_ptr[i].dns_name, (char)((122 * drand48()) + 97),
((sizeof(char) * 7) * drand48()) + (sizeof(char) * 2));
c random memset
c random memset
asked Nov 21 '18 at 19:19
jch3jch3
11
asked Nov 21 '18 at 19:19
jch3jch3
11
asked Nov 21 '18 at 19:19
jch3jch3
11
asked Nov 21 '18 at 19:19
jch3jch3
11
asked Nov 21 '18 at 19:19
jch3jch3
11
11
2
ASCII's range is only0-127. Unlessdrand48()returns0in(char)((122 * drand48()) + 97), it won't fit.
– Fiddling Bits
Nov 21 '18 at 19:20
1
Why are you forced to usememset? Doingdns_name[i] = ...seems more logical if you have to bang out multiple characters.
– tadman
Nov 21 '18 at 19:23
1
sizeof charis1per definition. Always. And1is neutral for multiplication, so drop it.lrand48()would be better fitting since you want integers.
– Swordfish
Nov 21 '18 at 19:24
2
. o O ( everytime one uses magic numbers, a kitty dies :( :( )
– Swordfish
Nov 21 '18 at 19:40
1
The problem withmemsetis that it's a very good way of setting multiple bytes to the same value, but a very poor way of setting many bytes to different values. In this case I think it would be far simpler to use one array assignment for each letter.
– Tim Randall
Nov 21 '18 at 19:59
|
show 7 more comments
2
ASCII's range is only0-127. Unlessdrand48()returns0in(char)((122 * drand48()) + 97), it won't fit.
– Fiddling Bits
Nov 21 '18 at 19:20
1
Why are you forced to usememset? Doingdns_name[i] = ...seems more logical if you have to bang out multiple characters.
– tadman
Nov 21 '18 at 19:23
1
sizeof charis1per definition. Always. And1is neutral for multiplication, so drop it.lrand48()would be better fitting since you want integers.
– Swordfish
Nov 21 '18 at 19:24
2
. o O ( everytime one uses magic numbers, a kitty dies :( :( )
– Swordfish
Nov 21 '18 at 19:40
1
The problem withmemsetis that it's a very good way of setting multiple bytes to the same value, but a very poor way of setting many bytes to different values. In this case I think it would be far simpler to use one array assignment for each letter.
– Tim Randall
Nov 21 '18 at 19:59
2
2
ASCII's range is only
0 - 127. Unless drand48() returns 0 in (char)((122 * drand48()) + 97), it won't fit.– Fiddling Bits
Nov 21 '18 at 19:20
ASCII's range is only
0 - 127. Unless drand48() returns 0 in (char)((122 * drand48()) + 97), it won't fit.– Fiddling Bits
Nov 21 '18 at 19:20
1
1
Why are you forced to use
memset? Doing dns_name[i] = ... seems more logical if you have to bang out multiple characters.– tadman
Nov 21 '18 at 19:23
Why are you forced to use
memset? Doing dns_name[i] = ... seems more logical if you have to bang out multiple characters.– tadman
Nov 21 '18 at 19:23
1
1
sizeof char is 1 per definition. Always. And 1 is neutral for multiplication, so drop it. lrand48() would be better fitting since you want integers.– Swordfish
Nov 21 '18 at 19:24
sizeof char is 1 per definition. Always. And 1 is neutral for multiplication, so drop it. lrand48() would be better fitting since you want integers.– Swordfish
Nov 21 '18 at 19:24
2
2
. o O ( everytime one uses magic numbers, a kitty dies :( :( )
– Swordfish
Nov 21 '18 at 19:40
. o O ( everytime one uses magic numbers, a kitty dies :( :( )
– Swordfish
Nov 21 '18 at 19:40
1
1
The problem with
memset is that it's a very good way of setting multiple bytes to the same value, but a very poor way of setting many bytes to different values. In this case I think it would be far simpler to use one array assignment for each letter.– Tim Randall
Nov 21 '18 at 19:59
The problem with
memset is that it's a very good way of setting multiple bytes to the same value, but a very poor way of setting many bytes to different values. In this case I think it would be far simpler to use one array assignment for each letter.– Tim Randall
Nov 21 '18 at 19:59
|
show 7 more comments
1 Answer
1
active
oldest
votes
My code returns non ASCII characters and I am not sure why.
Wrong scale used to generate lower case letters.
(122 * drand48()) + 97 converted to an integer type can readily make 122 different values. [97...218]. This is outside the ASCII range of [0...127].
How do I set a random number of random lowercase character ...
drand48() provides a random value [0...1.0). Scale by 26 and truncate to get 26 different indexes.
int index = (int) (drand48()*26); // 0...25
Pedantic code would be concerned about the few random values that may round the product to 26.0
if (index >= 26) index = 26 - 1;
int az = index + 'a';
// or look up in a table if non-ASCII encoding might be used
// 12345678901234567890123456
int az = "abcdefghijklmnopqrstuvwxyz"[index];
Selecting a random length would use the same thing, but with NUMLTRS instead of 26.
int length = (int) (drand48()*NUMLTRS);
if (index >= NUMLTRS) index = NUMLTRS -1;
... to a Struct member using memset in C
It is unclear if dns_name should be all the same, or generally different letters.
struct Record foo;
if (all_same) [
memset(foo.dns_name, az, length);
} else {
for (int i = 0; i < length; i++) {
int index = (int) (drand48()*26); // 0...25
if (index >= 26) index = 26 -1;
int az = index + 'a';
foo.dns_name[i] = az; // Does not make sense to use memset() here
}
}
Lastly, if dns_name is meant to be a string for ease of later use, declare with a +1 size
dns_name[NUMLTRS + 1];
// above code
foo.dns_name[NUMLTRS] = ''; // append null character
printf("dna_name <%s>n", foo.dns_name);
answered Nov 21 '18 at 22:57
chuxchux
84.8k874157
Can(int)(0.9999...)really get rounded up to1? "When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero)" (6.3.1.4 of ISO/IEC 9899:1999)
– usr2564301
Nov 21 '18 at 23:04
1
@usr2564301 The issue is not about(int)(0.9999...), but could0.9999... * 26under any rounding mode and anyFLT_EVAL_MODEresult in 26.0? True, mathematically0.9999... * 26is always less than 26, but this is FP math. A deep analysis may show 26.0 is not possible, but not having done that, a check is prudent.
– chux
Nov 21 '18 at 23:08
add a comment |
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My code returns non ASCII characters and I am not sure why.
Wrong scale used to generate lower case letters.
(122 * drand48()) + 97 converted to an integer type can readily make 122 different values. [97...218]. This is outside the ASCII range of [0...127].
How do I set a random number of random lowercase character ...
drand48() provides a random value [0...1.0). Scale by 26 and truncate to get 26 different indexes.
int index = (int) (drand48()*26); // 0...25
Pedantic code would be concerned about the few random values that may round the product to 26.0
if (index >= 26) index = 26 - 1;
int az = index + 'a';
// or look up in a table if non-ASCII encoding might be used
// 12345678901234567890123456
int az = "abcdefghijklmnopqrstuvwxyz"[index];
Selecting a random length would use the same thing, but with NUMLTRS instead of 26.
int length = (int) (drand48()*NUMLTRS);
if (index >= NUMLTRS) index = NUMLTRS -1;
... to a Struct member using memset in C
It is unclear if dns_name should be all the same, or generally different letters.
struct Record foo;
if (all_same) [
memset(foo.dns_name, az, length);
} else {
for (int i = 0; i < length; i++) {
int index = (int) (drand48()*26); // 0...25
if (index >= 26) index = 26 -1;
int az = index + 'a';
foo.dns_name[i] = az; // Does not make sense to use memset() here
}
}
Lastly, if dns_name is meant to be a string for ease of later use, declare with a +1 size
dns_name[NUMLTRS + 1];
// above code
foo.dns_name[NUMLTRS] = ''; // append null character
printf("dna_name <%s>n", foo.dns_name);
answered Nov 21 '18 at 22:57
chuxchux
84.8k874157
Can(int)(0.9999...)really get rounded up to1? "When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero)" (6.3.1.4 of ISO/IEC 9899:1999)
– usr2564301
Nov 21 '18 at 23:04
1
@usr2564301 The issue is not about(int)(0.9999...), but could0.9999... * 26under any rounding mode and anyFLT_EVAL_MODEresult in 26.0? True, mathematically0.9999... * 26is always less than 26, but this is FP math. A deep analysis may show 26.0 is not possible, but not having done that, a check is prudent.
– chux
Nov 21 '18 at 23:08
add a comment |
My code returns non ASCII characters and I am not sure why.
Wrong scale used to generate lower case letters.
(122 * drand48()) + 97 converted to an integer type can readily make 122 different values. [97...218]. This is outside the ASCII range of [0...127].
How do I set a random number of random lowercase character ...
drand48() provides a random value [0...1.0). Scale by 26 and truncate to get 26 different indexes.
int index = (int) (drand48()*26); // 0...25
Pedantic code would be concerned about the few random values that may round the product to 26.0
if (index >= 26) index = 26 - 1;
int az = index + 'a';
// or look up in a table if non-ASCII encoding might be used
// 12345678901234567890123456
int az = "abcdefghijklmnopqrstuvwxyz"[index];
Selecting a random length would use the same thing, but with NUMLTRS instead of 26.
int length = (int) (drand48()*NUMLTRS);
if (index >= NUMLTRS) index = NUMLTRS -1;
... to a Struct member using memset in C
It is unclear if dns_name should be all the same, or generally different letters.
struct Record foo;
if (all_same) [
memset(foo.dns_name, az, length);
} else {
for (int i = 0; i < length; i++) {
int index = (int) (drand48()*26); // 0...25
if (index >= 26) index = 26 -1;
int az = index + 'a';
foo.dns_name[i] = az; // Does not make sense to use memset() here
}
}
Lastly, if dns_name is meant to be a string for ease of later use, declare with a +1 size
dns_name[NUMLTRS + 1];
// above code
foo.dns_name[NUMLTRS] = ''; // append null character
printf("dna_name <%s>n", foo.dns_name);
answered Nov 21 '18 at 22:57
chuxchux
84.8k874157
Can(int)(0.9999...)really get rounded up to1? "When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero)" (6.3.1.4 of ISO/IEC 9899:1999)
– usr2564301
Nov 21 '18 at 23:04
1
@usr2564301 The issue is not about(int)(0.9999...), but could0.9999... * 26under any rounding mode and anyFLT_EVAL_MODEresult in 26.0? True, mathematically0.9999... * 26is always less than 26, but this is FP math. A deep analysis may show 26.0 is not possible, but not having done that, a check is prudent.
– chux
Nov 21 '18 at 23:08
add a comment |
My code returns non ASCII characters and I am not sure why.
Wrong scale used to generate lower case letters.
(122 * drand48()) + 97 converted to an integer type can readily make 122 different values. [97...218]. This is outside the ASCII range of [0...127].
How do I set a random number of random lowercase character ...
drand48() provides a random value [0...1.0). Scale by 26 and truncate to get 26 different indexes.
int index = (int) (drand48()*26); // 0...25
Pedantic code would be concerned about the few random values that may round the product to 26.0
if (index >= 26) index = 26 - 1;
int az = index + 'a';
// or look up in a table if non-ASCII encoding might be used
// 12345678901234567890123456
int az = "abcdefghijklmnopqrstuvwxyz"[index];
Selecting a random length would use the same thing, but with NUMLTRS instead of 26.
int length = (int) (drand48()*NUMLTRS);
if (index >= NUMLTRS) index = NUMLTRS -1;
... to a Struct member using memset in C
It is unclear if dns_name should be all the same, or generally different letters.
struct Record foo;
if (all_same) [
memset(foo.dns_name, az, length);
} else {
for (int i = 0; i < length; i++) {
int index = (int) (drand48()*26); // 0...25
if (index >= 26) index = 26 -1;
int az = index + 'a';
foo.dns_name[i] = az; // Does not make sense to use memset() here
}
}
Lastly, if dns_name is meant to be a string for ease of later use, declare with a +1 size
dns_name[NUMLTRS + 1];
// above code
foo.dns_name[NUMLTRS] = ''; // append null character
printf("dna_name <%s>n", foo.dns_name);
answered Nov 21 '18 at 22:57
chuxchux
84.8k874157
My code returns non ASCII characters and I am not sure why.
Wrong scale used to generate lower case letters.
(122 * drand48()) + 97 converted to an integer type can readily make 122 different values. [97...218]. This is outside the ASCII range of [0...127].
How do I set a random number of random lowercase character ...
drand48() provides a random value [0...1.0). Scale by 26 and truncate to get 26 different indexes.
int index = (int) (drand48()*26); // 0...25
Pedantic code would be concerned about the few random values that may round the product to 26.0
if (index >= 26) index = 26 - 1;
int az = index + 'a';
// or look up in a table if non-ASCII encoding might be used
// 12345678901234567890123456
int az = "abcdefghijklmnopqrstuvwxyz"[index];
Selecting a random length would use the same thing, but with NUMLTRS instead of 26.
int length = (int) (drand48()*NUMLTRS);
if (index >= NUMLTRS) index = NUMLTRS -1;
... to a Struct member using memset in C
It is unclear if dns_name should be all the same, or generally different letters.
struct Record foo;
if (all_same) [
memset(foo.dns_name, az, length);
} else {
for (int i = 0; i < length; i++) {
int index = (int) (drand48()*26); // 0...25
if (index >= 26) index = 26 -1;
int az = index + 'a';
foo.dns_name[i] = az; // Does not make sense to use memset() here
}
}
Lastly, if dns_name is meant to be a string for ease of later use, declare with a +1 size
dns_name[NUMLTRS + 1];
// above code
foo.dns_name[NUMLTRS] = ''; // append null character
printf("dna_name <%s>n", foo.dns_name);
answered Nov 21 '18 at 22:57
chuxchux
84.8k874157
edited Nov 21 '18 at 23:20
answered Nov 21 '18 at 22:57
chuxchux
84.8k874157
answered Nov 21 '18 at 22:57
chuxchux
84.8k874157
answered Nov 21 '18 at 22:57
chuxchux
84.8k874157
84.8k874157
Can(int)(0.9999...)really get rounded up to1? "When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero)" (6.3.1.4 of ISO/IEC 9899:1999)
– usr2564301
Nov 21 '18 at 23:04
1
@usr2564301 The issue is not about(int)(0.9999...), but could0.9999... * 26under any rounding mode and anyFLT_EVAL_MODEresult in 26.0? True, mathematically0.9999... * 26is always less than 26, but this is FP math. A deep analysis may show 26.0 is not possible, but not having done that, a check is prudent.
– chux
Nov 21 '18 at 23:08
add a comment |
Can(int)(0.9999...)really get rounded up to1? "When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero)" (6.3.1.4 of ISO/IEC 9899:1999)
– usr2564301
Nov 21 '18 at 23:04
1
@usr2564301 The issue is not about(int)(0.9999...), but could0.9999... * 26under any rounding mode and anyFLT_EVAL_MODEresult in 26.0? True, mathematically0.9999... * 26is always less than 26, but this is FP math. A deep analysis may show 26.0 is not possible, but not having done that, a check is prudent.
– chux
Nov 21 '18 at 23:08
Can
(int)(0.9999...) really get rounded up to 1? "When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero)" (6.3.1.4 of ISO/IEC 9899:1999)– usr2564301
Nov 21 '18 at 23:04
Can
(int)(0.9999...) really get rounded up to 1? "When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero)" (6.3.1.4 of ISO/IEC 9899:1999)– usr2564301
Nov 21 '18 at 23:04
1
1
@usr2564301 The issue is not about
(int)(0.9999...), but could 0.9999... * 26 under any rounding mode and any FLT_EVAL_MODE result in 26.0? True, mathematically 0.9999... * 26 is always less than 26, but this is FP math. A deep analysis may show 26.0 is not possible, but not having done that, a check is prudent.– chux
Nov 21 '18 at 23:08
@usr2564301 The issue is not about
(int)(0.9999...), but could 0.9999... * 26 under any rounding mode and any FLT_EVAL_MODE result in 26.0? True, mathematically 0.9999... * 26 is always less than 26, but this is FP math. A deep analysis may show 26.0 is not possible, but not having done that, a check is prudent.– chux
Nov 21 '18 at 23:08
add a comment |
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2
ASCII's range is only
0-127. Unlessdrand48()returns0in(char)((122 * drand48()) + 97), it won't fit.– Fiddling Bits
Nov 21 '18 at 19:20
1
Why are you forced to use
memset? Doingdns_name[i] = ...seems more logical if you have to bang out multiple characters.– tadman
Nov 21 '18 at 19:23
1
sizeof charis1per definition. Always. And1is neutral for multiplication, so drop it.lrand48()would be better fitting since you want integers.– Swordfish
Nov 21 '18 at 19:24
2
. o O ( everytime one uses magic numbers, a kitty dies :( :( )
– Swordfish
Nov 21 '18 at 19:40
1
The problem with
memsetis that it's a very good way of setting multiple bytes to the same value, but a very poor way of setting many bytes to different values. In this case I think it would be far simpler to use one array assignment for each letter.– Tim Randall
Nov 21 '18 at 19:59