Nested for-loop: giving concatenated result
I am currently finding it hard to grasp the output from a simple nested for-loop
int result = 0;
for (int i = 1; i <= 3; i++)
{
for (int j = 1; j <= 2; j++)
{
result = result +i;
}
}
Console.WriteLine(result);
Which gives the output of 12. I understand that the result for j is 1 and 2, but just cannot grasp why the final result is 12 instead of 3?
Can someone explain please?
c#
edited Nov 19 '18 at 6:37
Cœur
17.8k9106145
asked Jun 10 '18 at 7:50
NZ_DJNZ_DJ
12010
add a comment |
I am currently finding it hard to grasp the output from a simple nested for-loop
int result = 0;
for (int i = 1; i <= 3; i++)
{
for (int j = 1; j <= 2; j++)
{
result = result +i;
}
}
Console.WriteLine(result);
Which gives the output of 12. I understand that the result for j is 1 and 2, but just cannot grasp why the final result is 12 instead of 3?
Can someone explain please?
c#
edited Nov 19 '18 at 6:37
Cœur
17.8k9106145
asked Jun 10 '18 at 7:50
NZ_DJNZ_DJ
12010
Step through it in your debugger. Follow the logic step by step, line by line, iteration by iteration. Then you'll know.
– Anthony Pegram
Jun 10 '18 at 7:53
0+1+1+2+2+6+6=12. Think about what your loop does and add more outputs in between to show the current value of result..
– ArgusMagnus
Jun 10 '18 at 7:55
add a comment |
I am currently finding it hard to grasp the output from a simple nested for-loop
int result = 0;
for (int i = 1; i <= 3; i++)
{
for (int j = 1; j <= 2; j++)
{
result = result +i;
}
}
Console.WriteLine(result);
Which gives the output of 12. I understand that the result for j is 1 and 2, but just cannot grasp why the final result is 12 instead of 3?
Can someone explain please?
c#
edited Nov 19 '18 at 6:37
Cœur
17.8k9106145
asked Jun 10 '18 at 7:50
NZ_DJNZ_DJ
12010
I am currently finding it hard to grasp the output from a simple nested for-loop
int result = 0;
for (int i = 1; i <= 3; i++)
{
for (int j = 1; j <= 2; j++)
{
result = result +i;
}
}
Console.WriteLine(result);
Which gives the output of 12. I understand that the result for j is 1 and 2, but just cannot grasp why the final result is 12 instead of 3?
Can someone explain please?
c#
c#
edited Nov 19 '18 at 6:37
Cœur
17.8k9106145
asked Jun 10 '18 at 7:50
NZ_DJNZ_DJ
12010
edited Nov 19 '18 at 6:37
Cœur
17.8k9106145
asked Jun 10 '18 at 7:50
NZ_DJNZ_DJ
12010
edited Nov 19 '18 at 6:37
Cœur
17.8k9106145
edited Nov 19 '18 at 6:37
Cœur
17.8k9106145
edited Nov 19 '18 at 6:37
Cœur
17.8k9106145
17.8k9106145
asked Jun 10 '18 at 7:50
NZ_DJNZ_DJ
12010
asked Jun 10 '18 at 7:50
NZ_DJNZ_DJ
12010
asked Jun 10 '18 at 7:50
NZ_DJNZ_DJ
12010
12010
Step through it in your debugger. Follow the logic step by step, line by line, iteration by iteration. Then you'll know.
– Anthony Pegram
Jun 10 '18 at 7:53
0+1+1+2+2+6+6=12. Think about what your loop does and add more outputs in between to show the current value of result..
– ArgusMagnus
Jun 10 '18 at 7:55
add a comment |
Step through it in your debugger. Follow the logic step by step, line by line, iteration by iteration. Then you'll know.
– Anthony Pegram
Jun 10 '18 at 7:53
0+1+1+2+2+6+6=12. Think about what your loop does and add more outputs in between to show the current value of result..
– ArgusMagnus
Jun 10 '18 at 7:55
Step through it in your debugger. Follow the logic step by step, line by line, iteration by iteration. Then you'll know.
– Anthony Pegram
Jun 10 '18 at 7:53
Step through it in your debugger. Follow the logic step by step, line by line, iteration by iteration. Then you'll know.
– Anthony Pegram
Jun 10 '18 at 7:53
0+1+1+2+2+6+6=12. Think about what your loop does and add more outputs in between to show the current value of result..
– ArgusMagnus
Jun 10 '18 at 7:55
0+1+1+2+2+6+6=12. Think about what your loop does and add more outputs in between to show the current value of result..
– ArgusMagnus
Jun 10 '18 at 7:55
add a comment |
3 Answers
3
active
oldest
votes
Your outer loop has three iterations. On each of those three iterations, you're running the inner loop:
for (int j = 1; j <= 2; j++)
{
result = result +i;
}
Given that the body of the loop doesn't depend on j, and the inner loop will always run twice, we can replace the whole of the inner loop with:
result = result + i;
result = result + i;
Or to simplify it:
result = result + (i * 2);
So now the whole code is equivalent to:
int result = 0;
for (int i = 1; i <= 3; i++)
{
result = result + (i * 2);
}
Console.WriteLine(result);
So after the first iteration of the outer loop, result is 2 (we've added 2 to 0).
After the second iteration of the outer loop, result is 6 (we've added 4 to 2).
After the third iteration of the outer loop, result is 12 (we've added 6 to 6).
answered Jun 10 '18 at 7:54
Jon SkeetJon Skeet
1083k68279098417
Thanks mate, what was I thinking just putting 1 and 2 together duh, Cheers!
– NZ_DJ
Jun 10 '18 at 7:58
add a comment |
You are adding the value of i to result each iteration of the inner loop, and the inner loop runs 3 times (the 3 iterations of the outer loop). Therefore the first time the inner loop runs, you add 1 twice, the second time the inner loop runs you add 2 twice, and the final time it runs you add 3 twice:
result = 1 + 1 + 2 + 2 + 3 + 3 = 12
answered Jun 10 '18 at 7:53
IridiumIridium
17.3k34056
add a comment |
run that test and you will see:
class Program
{
public static void test()
{
int result = 0;
for (int i = 1; i <= 3; i++)
{
Console.WriteLine("i["+i+"] : ");
for (int j = 1; j <= 2; j++)
{
Console.Write(" when : j["+j+"]");
Console.WriteLine("Adding " + i + "to" + result);
result = result + i;
}
}
Console.WriteLine(result);
Console.ReadLine();
}
static void Main(string args)
{
test();
}
}
answered Jun 10 '18 at 8:03
Leon BarkanLeon Barkan
2,04321129
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your outer loop has three iterations. On each of those three iterations, you're running the inner loop:
for (int j = 1; j <= 2; j++)
{
result = result +i;
}
Given that the body of the loop doesn't depend on j, and the inner loop will always run twice, we can replace the whole of the inner loop with:
result = result + i;
result = result + i;
Or to simplify it:
result = result + (i * 2);
So now the whole code is equivalent to:
int result = 0;
for (int i = 1; i <= 3; i++)
{
result = result + (i * 2);
}
Console.WriteLine(result);
So after the first iteration of the outer loop, result is 2 (we've added 2 to 0).
After the second iteration of the outer loop, result is 6 (we've added 4 to 2).
After the third iteration of the outer loop, result is 12 (we've added 6 to 6).
answered Jun 10 '18 at 7:54
Jon SkeetJon Skeet
1083k68279098417
Thanks mate, what was I thinking just putting 1 and 2 together duh, Cheers!
– NZ_DJ
Jun 10 '18 at 7:58
add a comment |
Your outer loop has three iterations. On each of those three iterations, you're running the inner loop:
for (int j = 1; j <= 2; j++)
{
result = result +i;
}
Given that the body of the loop doesn't depend on j, and the inner loop will always run twice, we can replace the whole of the inner loop with:
result = result + i;
result = result + i;
Or to simplify it:
result = result + (i * 2);
So now the whole code is equivalent to:
int result = 0;
for (int i = 1; i <= 3; i++)
{
result = result + (i * 2);
}
Console.WriteLine(result);
So after the first iteration of the outer loop, result is 2 (we've added 2 to 0).
After the second iteration of the outer loop, result is 6 (we've added 4 to 2).
After the third iteration of the outer loop, result is 12 (we've added 6 to 6).
answered Jun 10 '18 at 7:54
Jon SkeetJon Skeet
1083k68279098417
Thanks mate, what was I thinking just putting 1 and 2 together duh, Cheers!
– NZ_DJ
Jun 10 '18 at 7:58
add a comment |
Your outer loop has three iterations. On each of those three iterations, you're running the inner loop:
for (int j = 1; j <= 2; j++)
{
result = result +i;
}
Given that the body of the loop doesn't depend on j, and the inner loop will always run twice, we can replace the whole of the inner loop with:
result = result + i;
result = result + i;
Or to simplify it:
result = result + (i * 2);
So now the whole code is equivalent to:
int result = 0;
for (int i = 1; i <= 3; i++)
{
result = result + (i * 2);
}
Console.WriteLine(result);
So after the first iteration of the outer loop, result is 2 (we've added 2 to 0).
After the second iteration of the outer loop, result is 6 (we've added 4 to 2).
After the third iteration of the outer loop, result is 12 (we've added 6 to 6).
answered Jun 10 '18 at 7:54
Jon SkeetJon Skeet
1083k68279098417
Your outer loop has three iterations. On each of those three iterations, you're running the inner loop:
for (int j = 1; j <= 2; j++)
{
result = result +i;
}
Given that the body of the loop doesn't depend on j, and the inner loop will always run twice, we can replace the whole of the inner loop with:
result = result + i;
result = result + i;
Or to simplify it:
result = result + (i * 2);
So now the whole code is equivalent to:
int result = 0;
for (int i = 1; i <= 3; i++)
{
result = result + (i * 2);
}
Console.WriteLine(result);
So after the first iteration of the outer loop, result is 2 (we've added 2 to 0).
After the second iteration of the outer loop, result is 6 (we've added 4 to 2).
After the third iteration of the outer loop, result is 12 (we've added 6 to 6).
answered Jun 10 '18 at 7:54
Jon SkeetJon Skeet
1083k68279098417
answered Jun 10 '18 at 7:54
Jon SkeetJon Skeet
1083k68279098417
answered Jun 10 '18 at 7:54
Jon SkeetJon Skeet
1083k68279098417
answered Jun 10 '18 at 7:54
Jon SkeetJon Skeet
1083k68279098417
1083k68279098417
Thanks mate, what was I thinking just putting 1 and 2 together duh, Cheers!
– NZ_DJ
Jun 10 '18 at 7:58
add a comment |
Thanks mate, what was I thinking just putting 1 and 2 together duh, Cheers!
– NZ_DJ
Jun 10 '18 at 7:58
Thanks mate, what was I thinking just putting 1 and 2 together duh, Cheers!
– NZ_DJ
Jun 10 '18 at 7:58
Thanks mate, what was I thinking just putting 1 and 2 together duh, Cheers!
– NZ_DJ
Jun 10 '18 at 7:58
add a comment |
You are adding the value of i to result each iteration of the inner loop, and the inner loop runs 3 times (the 3 iterations of the outer loop). Therefore the first time the inner loop runs, you add 1 twice, the second time the inner loop runs you add 2 twice, and the final time it runs you add 3 twice:
result = 1 + 1 + 2 + 2 + 3 + 3 = 12
answered Jun 10 '18 at 7:53
IridiumIridium
17.3k34056
add a comment |
You are adding the value of i to result each iteration of the inner loop, and the inner loop runs 3 times (the 3 iterations of the outer loop). Therefore the first time the inner loop runs, you add 1 twice, the second time the inner loop runs you add 2 twice, and the final time it runs you add 3 twice:
result = 1 + 1 + 2 + 2 + 3 + 3 = 12
answered Jun 10 '18 at 7:53
IridiumIridium
17.3k34056
add a comment |
You are adding the value of i to result each iteration of the inner loop, and the inner loop runs 3 times (the 3 iterations of the outer loop). Therefore the first time the inner loop runs, you add 1 twice, the second time the inner loop runs you add 2 twice, and the final time it runs you add 3 twice:
result = 1 + 1 + 2 + 2 + 3 + 3 = 12
answered Jun 10 '18 at 7:53
IridiumIridium
17.3k34056
You are adding the value of i to result each iteration of the inner loop, and the inner loop runs 3 times (the 3 iterations of the outer loop). Therefore the first time the inner loop runs, you add 1 twice, the second time the inner loop runs you add 2 twice, and the final time it runs you add 3 twice:
result = 1 + 1 + 2 + 2 + 3 + 3 = 12
answered Jun 10 '18 at 7:53
IridiumIridium
17.3k34056
answered Jun 10 '18 at 7:53
IridiumIridium
17.3k34056
answered Jun 10 '18 at 7:53
IridiumIridium
17.3k34056
answered Jun 10 '18 at 7:53
IridiumIridium
17.3k34056
17.3k34056
add a comment |
add a comment |
run that test and you will see:
class Program
{
public static void test()
{
int result = 0;
for (int i = 1; i <= 3; i++)
{
Console.WriteLine("i["+i+"] : ");
for (int j = 1; j <= 2; j++)
{
Console.Write(" when : j["+j+"]");
Console.WriteLine("Adding " + i + "to" + result);
result = result + i;
}
}
Console.WriteLine(result);
Console.ReadLine();
}
static void Main(string args)
{
test();
}
}
answered Jun 10 '18 at 8:03
Leon BarkanLeon Barkan
2,04321129
add a comment |
run that test and you will see:
class Program
{
public static void test()
{
int result = 0;
for (int i = 1; i <= 3; i++)
{
Console.WriteLine("i["+i+"] : ");
for (int j = 1; j <= 2; j++)
{
Console.Write(" when : j["+j+"]");
Console.WriteLine("Adding " + i + "to" + result);
result = result + i;
}
}
Console.WriteLine(result);
Console.ReadLine();
}
static void Main(string args)
{
test();
}
}
answered Jun 10 '18 at 8:03
Leon BarkanLeon Barkan
2,04321129
add a comment |
run that test and you will see:
class Program
{
public static void test()
{
int result = 0;
for (int i = 1; i <= 3; i++)
{
Console.WriteLine("i["+i+"] : ");
for (int j = 1; j <= 2; j++)
{
Console.Write(" when : j["+j+"]");
Console.WriteLine("Adding " + i + "to" + result);
result = result + i;
}
}
Console.WriteLine(result);
Console.ReadLine();
}
static void Main(string args)
{
test();
}
}
answered Jun 10 '18 at 8:03
Leon BarkanLeon Barkan
2,04321129
run that test and you will see:
class Program
{
public static void test()
{
int result = 0;
for (int i = 1; i <= 3; i++)
{
Console.WriteLine("i["+i+"] : ");
for (int j = 1; j <= 2; j++)
{
Console.Write(" when : j["+j+"]");
Console.WriteLine("Adding " + i + "to" + result);
result = result + i;
}
}
Console.WriteLine(result);
Console.ReadLine();
}
static void Main(string args)
{
test();
}
}
answered Jun 10 '18 at 8:03
Leon BarkanLeon Barkan
2,04321129
answered Jun 10 '18 at 8:03
Leon BarkanLeon Barkan
2,04321129
answered Jun 10 '18 at 8:03
Leon BarkanLeon Barkan
2,04321129
answered Jun 10 '18 at 8:03
Leon BarkanLeon Barkan
2,04321129
2,04321129
add a comment |
add a comment |
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Step through it in your debugger. Follow the logic step by step, line by line, iteration by iteration. Then you'll know.
– Anthony Pegram
Jun 10 '18 at 7:53
0+1+1+2+2+6+6=12. Think about what your loop does and add more outputs in between to show the current value of result..
– ArgusMagnus
Jun 10 '18 at 7:55