C: how to create an empty STRING?

Please I have been struggling for 5 hours on this, I know how to create a empty char it's like this :

char c = '';

but how can I create en empty STRING ? (because I want to use it as a temporary place to store other strings)

char temporary = "";

error : array type 'char [1]' is not assignable

edited Nov 20 '18 at 18:49

asked Nov 20 '18 at 18:18

masterCUCK9

2

Your question is unclear. What do you mean by an "empty char"? In any event, it is not possible in C to magically create something that can be used to contain an arbitrary and changeable string of arbitrary length, unless you dynamically allocate and manage memory. Also, the second example (in the form you've shown it) is valid - if you're getting an error from that, your code differs in some way. Consider providing an actual sample of actual code in the form of a Minimal, Complete, and Verifiable example.

– Peter
Nov 20 '18 at 18:29

tempo = ""; Assuming tempo is a char array, you can't do that. Try tempo[0] = ''. You could also do: strcpy(temp, ""); Or to completely clear it: memset(tempo, 0, sizeof(tempo);. But that's probably not necessary.

– Johnny Mopp
Nov 20 '18 at 18:32

can I do strcpy(temp, ""); without having said what temp is ???

– masterCUCK9
Nov 20 '18 at 18:38

I can't use strcpy if temp is not DEFINED

– masterCUCK9
Nov 20 '18 at 18:39

@masterCUCK9 A typo. I meant to type tempo. As in: strcpy(tempo, "");

– Johnny Mopp
Nov 20 '18 at 18:39

|
show 8 more comments

Please I have been struggling for 5 hours on this, I know how to create a empty char it's like this :

char c = '';

but how can I create en empty STRING ? (because I want to use it as a temporary place to store other strings)

char temporary = "";

error : array type 'char [1]' is not assignable

edited Nov 20 '18 at 18:49

asked Nov 20 '18 at 18:18

masterCUCK9

2

Your question is unclear. What do you mean by an "empty char"? In any event, it is not possible in C to magically create something that can be used to contain an arbitrary and changeable string of arbitrary length, unless you dynamically allocate and manage memory. Also, the second example (in the form you've shown it) is valid - if you're getting an error from that, your code differs in some way. Consider providing an actual sample of actual code in the form of a Minimal, Complete, and Verifiable example.

– Peter
Nov 20 '18 at 18:29

tempo = ""; Assuming tempo is a char array, you can't do that. Try tempo[0] = ''. You could also do: strcpy(temp, ""); Or to completely clear it: memset(tempo, 0, sizeof(tempo);. But that's probably not necessary.

– Johnny Mopp
Nov 20 '18 at 18:32

can I do strcpy(temp, ""); without having said what temp is ???

– masterCUCK9
Nov 20 '18 at 18:38

I can't use strcpy if temp is not DEFINED

– masterCUCK9
Nov 20 '18 at 18:39

@masterCUCK9 A typo. I meant to type tempo. As in: strcpy(tempo, "");

– Johnny Mopp
Nov 20 '18 at 18:39

|
show 8 more comments

Please I have been struggling for 5 hours on this, I know how to create a empty char it's like this :

char c = '';

but how can I create en empty STRING ? (because I want to use it as a temporary place to store other strings)

char temporary = "";

error : array type 'char [1]' is not assignable

edited Nov 20 '18 at 18:49

asked Nov 20 '18 at 18:18

masterCUCK9

Please I have been struggling for 5 hours on this, I know how to create a empty char it's like this :

char c = '';

but how can I create en empty STRING ? (because I want to use it as a temporary place to store other strings)

char temporary = "";

error : array type 'char [1]' is not assignable

c arrays string

edited Nov 20 '18 at 18:49

asked Nov 20 '18 at 18:18

masterCUCK9

edited Nov 20 '18 at 18:49

asked Nov 20 '18 at 18:18

masterCUCK9

edited Nov 20 '18 at 18:49

asked Nov 20 '18 at 18:18

masterCUCK9

asked Nov 20 '18 at 18:18

masterCUCK9

asked Nov 20 '18 at 18:18

masterCUCK9

2

Your question is unclear. What do you mean by an "empty char"? In any event, it is not possible in C to magically create something that can be used to contain an arbitrary and changeable string of arbitrary length, unless you dynamically allocate and manage memory. Also, the second example (in the form you've shown it) is valid - if you're getting an error from that, your code differs in some way. Consider providing an actual sample of actual code in the form of a Minimal, Complete, and Verifiable example.

– Peter
Nov 20 '18 at 18:29

tempo = ""; Assuming tempo is a char array, you can't do that. Try tempo[0] = ''. You could also do: strcpy(temp, ""); Or to completely clear it: memset(tempo, 0, sizeof(tempo);. But that's probably not necessary.

– Johnny Mopp
Nov 20 '18 at 18:32

can I do strcpy(temp, ""); without having said what temp is ???

– masterCUCK9
Nov 20 '18 at 18:38

I can't use strcpy if temp is not DEFINED

– masterCUCK9
Nov 20 '18 at 18:39

@masterCUCK9 A typo. I meant to type tempo. As in: strcpy(tempo, "");

– Johnny Mopp
Nov 20 '18 at 18:39

|
show 8 more comments

2

Your question is unclear. What do you mean by an "empty char"? In any event, it is not possible in C to magically create something that can be used to contain an arbitrary and changeable string of arbitrary length, unless you dynamically allocate and manage memory. Also, the second example (in the form you've shown it) is valid - if you're getting an error from that, your code differs in some way. Consider providing an actual sample of actual code in the form of a Minimal, Complete, and Verifiable example.

– Peter
Nov 20 '18 at 18:29

tempo = ""; Assuming tempo is a char array, you can't do that. Try tempo[0] = ''. You could also do: strcpy(temp, ""); Or to completely clear it: memset(tempo, 0, sizeof(tempo);. But that's probably not necessary.

– Johnny Mopp
Nov 20 '18 at 18:32

can I do strcpy(temp, ""); without having said what temp is ???

– masterCUCK9
Nov 20 '18 at 18:38

I can't use strcpy if temp is not DEFINED

– masterCUCK9
Nov 20 '18 at 18:39

@masterCUCK9 A typo. I meant to type tempo. As in: strcpy(tempo, "");

– Johnny Mopp
Nov 20 '18 at 18:39

Your question is unclear. What do you mean by an "empty char"? In any event, it is not possible in C to magically create something that can be used to contain an arbitrary and changeable string of arbitrary length, unless you dynamically allocate and manage memory. Also, the second example (in the form you've shown it) is valid - if you're getting an error from that, your code differs in some way. Consider providing an actual sample of actual code in the form of a Minimal, Complete, and Verifiable example.

– Peter
Nov 20 '18 at 18:29

tempo = ""; Assuming tempo is a char array, you can't do that. Try tempo[0] = ''. You could also do: strcpy(temp, ""); Or to completely clear it: memset(tempo, 0, sizeof(tempo);. But that's probably not necessary.

– Johnny Mopp
Nov 20 '18 at 18:32

can I do strcpy(temp, ""); without having said what temp is ???

– masterCUCK9
Nov 20 '18 at 18:38

I can't use strcpy if temp is not DEFINED

– masterCUCK9
Nov 20 '18 at 18:39

@masterCUCK9 A typo. I meant to type tempo. As in: strcpy(tempo, "");

– Johnny Mopp
Nov 20 '18 at 18:39

|
show 8 more comments

2 Answers
2

active

oldest

votes

Define a buffer to hold the string contents:

#define BUFFER_SIZE 256 // or whatever size you need



char buffer[BUFFER_SIZE+1] = {0}; // +1 for string terminator, 

                                  // = {0} initializer zeroes out entire buffer

To assign a string to this buffer, use strcpy:

strcpy( buffer, "some string" );

To append a string to this buffer, use strcat:

strcat( buffer, "more string" );

EDIT

Now that you've edited your question, the problem is the line

tempo = "";

An array expression like tempo may not be the target of the = operator; you must use a library function like strcpy to assign string values. If you want to set tempo to an empty string, you can do any one of the following:

strcpy( tempo, "" );

tempo[0] = 0;

tempo[0] = '';

edited Nov 20 '18 at 18:36

answered Nov 20 '18 at 18:30

John Bode

83k1378150

I would actually memset tempo to its size as 0's to the entire buffer is reset instead of just the first character.

– ritlew
Nov 20 '18 at 18:38

1

@ritlew: It kind of depends on what the OP is wanting to do - if all he needs to do is make sure tempo contains an empty string before a strcpy operation, then setting the initial element to 0 is sufficient. If he's writing this buffer to a file or something, then yes, it would be safer to zero out all the unused elements.

– John Bode
Nov 20 '18 at 18:39

Well if the string is "Hello", then it becomes "ello". So if they were to strncpy 2 characters "Hi" into the buffer, it becomes "Hillo".

– ritlew
Nov 20 '18 at 18:42

AAAAYYYAAA I FOUND OUT THE PROBLEM

– masterCUCK9
Nov 20 '18 at 18:48

@ritlew: And this is one of the reasons I never use strncpy.

– John Bode
Nov 20 '18 at 20:11

add a comment |

In C you have to create a temporary buffer all the time. The most typical way to do this is to write a line of code like this:

char temp_buffer[256];

This buffer will have 256 characters of space available to be able to put strings in.

edited Nov 20 '18 at 18:50

answered Nov 20 '18 at 18:22

ritlew

853411

Comments are not for extended discussion; this conversation has been moved to chat.

– Samuel Liew♦
Nov 21 '18 at 2:15

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Define a buffer to hold the string contents:

#define BUFFER_SIZE 256 // or whatever size you need



char buffer[BUFFER_SIZE+1] = {0}; // +1 for string terminator, 

                                  // = {0} initializer zeroes out entire buffer

To assign a string to this buffer, use strcpy:

strcpy( buffer, "some string" );

To append a string to this buffer, use strcat:

strcat( buffer, "more string" );

EDIT

Now that you've edited your question, the problem is the line

tempo = "";

strcpy( tempo, "" );

tempo[0] = 0;

tempo[0] = '';

edited Nov 20 '18 at 18:36

answered Nov 20 '18 at 18:30

John Bode

83k1378150

I would actually memset tempo to its size as 0's to the entire buffer is reset instead of just the first character.

– ritlew
Nov 20 '18 at 18:38

1

@ritlew: It kind of depends on what the OP is wanting to do - if all he needs to do is make sure tempo contains an empty string before a strcpy operation, then setting the initial element to 0 is sufficient. If he's writing this buffer to a file or something, then yes, it would be safer to zero out all the unused elements.

– John Bode
Nov 20 '18 at 18:39

Well if the string is "Hello", then it becomes "ello". So if they were to strncpy 2 characters "Hi" into the buffer, it becomes "Hillo".

– ritlew
Nov 20 '18 at 18:42

AAAAYYYAAA I FOUND OUT THE PROBLEM

– masterCUCK9
Nov 20 '18 at 18:48

@ritlew: And this is one of the reasons I never use strncpy.

– John Bode
Nov 20 '18 at 20:11

add a comment |

Define a buffer to hold the string contents:

#define BUFFER_SIZE 256 // or whatever size you need



char buffer[BUFFER_SIZE+1] = {0}; // +1 for string terminator, 

                                  // = {0} initializer zeroes out entire buffer

To assign a string to this buffer, use strcpy:

strcpy( buffer, "some string" );

To append a string to this buffer, use strcat:

strcat( buffer, "more string" );

EDIT

Now that you've edited your question, the problem is the line

tempo = "";

strcpy( tempo, "" );

tempo[0] = 0;

tempo[0] = '';

edited Nov 20 '18 at 18:36

answered Nov 20 '18 at 18:30

John Bode

83k1378150

I would actually memset tempo to its size as 0's to the entire buffer is reset instead of just the first character.

– ritlew
Nov 20 '18 at 18:38

1

@ritlew: It kind of depends on what the OP is wanting to do - if all he needs to do is make sure tempo contains an empty string before a strcpy operation, then setting the initial element to 0 is sufficient. If he's writing this buffer to a file or something, then yes, it would be safer to zero out all the unused elements.

– John Bode
Nov 20 '18 at 18:39

Well if the string is "Hello", then it becomes "ello". So if they were to strncpy 2 characters "Hi" into the buffer, it becomes "Hillo".

– ritlew
Nov 20 '18 at 18:42

AAAAYYYAAA I FOUND OUT THE PROBLEM

– masterCUCK9
Nov 20 '18 at 18:48

@ritlew: And this is one of the reasons I never use strncpy.

– John Bode
Nov 20 '18 at 20:11

add a comment |

Define a buffer to hold the string contents:

#define BUFFER_SIZE 256 // or whatever size you need



char buffer[BUFFER_SIZE+1] = {0}; // +1 for string terminator, 

                                  // = {0} initializer zeroes out entire buffer

To assign a string to this buffer, use strcpy:

strcpy( buffer, "some string" );

To append a string to this buffer, use strcat:

strcat( buffer, "more string" );

EDIT

Now that you've edited your question, the problem is the line

tempo = "";

strcpy( tempo, "" );

tempo[0] = 0;

tempo[0] = '';

edited Nov 20 '18 at 18:36

answered Nov 20 '18 at 18:30

John Bode

83k1378150

Define a buffer to hold the string contents:

#define BUFFER_SIZE 256 // or whatever size you need



char buffer[BUFFER_SIZE+1] = {0}; // +1 for string terminator, 

                                  // = {0} initializer zeroes out entire buffer

To assign a string to this buffer, use strcpy:

strcpy( buffer, "some string" );

To append a string to this buffer, use strcat:

strcat( buffer, "more string" );

EDIT

Now that you've edited your question, the problem is the line

tempo = "";

strcpy( tempo, "" );

tempo[0] = 0;

tempo[0] = '';

edited Nov 20 '18 at 18:36

answered Nov 20 '18 at 18:30

John Bode

83k1378150

edited Nov 20 '18 at 18:36

answered Nov 20 '18 at 18:30

John Bode

83k1378150

answered Nov 20 '18 at 18:30

John Bode

83k1378150

answered Nov 20 '18 at 18:30

John Bode

83k1378150

I would actually memset tempo to its size as 0's to the entire buffer is reset instead of just the first character.

– ritlew
Nov 20 '18 at 18:38

1

@ritlew: It kind of depends on what the OP is wanting to do - if all he needs to do is make sure tempo contains an empty string before a strcpy operation, then setting the initial element to 0 is sufficient. If he's writing this buffer to a file or something, then yes, it would be safer to zero out all the unused elements.

– John Bode
Nov 20 '18 at 18:39

Well if the string is "Hello", then it becomes "ello". So if they were to strncpy 2 characters "Hi" into the buffer, it becomes "Hillo".

– ritlew
Nov 20 '18 at 18:42

AAAAYYYAAA I FOUND OUT THE PROBLEM

– masterCUCK9
Nov 20 '18 at 18:48

@ritlew: And this is one of the reasons I never use strncpy.

– John Bode
Nov 20 '18 at 20:11

add a comment |

I would actually memset tempo to its size as 0's to the entire buffer is reset instead of just the first character.

– ritlew
Nov 20 '18 at 18:38

1

@ritlew: It kind of depends on what the OP is wanting to do - if all he needs to do is make sure tempo contains an empty string before a strcpy operation, then setting the initial element to 0 is sufficient. If he's writing this buffer to a file or something, then yes, it would be safer to zero out all the unused elements.

– John Bode
Nov 20 '18 at 18:39

Well if the string is "Hello", then it becomes "ello". So if they were to strncpy 2 characters "Hi" into the buffer, it becomes "Hillo".

– ritlew
Nov 20 '18 at 18:42

AAAAYYYAAA I FOUND OUT THE PROBLEM

– masterCUCK9
Nov 20 '18 at 18:48

@ritlew: And this is one of the reasons I never use strncpy.

– John Bode
Nov 20 '18 at 20:11

I would actually memset tempo to its size as 0's to the entire buffer is reset instead of just the first character.

– ritlew
Nov 20 '18 at 18:38

@ritlew: It kind of depends on what the OP is wanting to do - if all he needs to do is make sure tempo contains an empty string before a strcpy operation, then setting the initial element to 0 is sufficient. If he's writing this buffer to a file or something, then yes, it would be safer to zero out all the unused elements.

– John Bode
Nov 20 '18 at 18:39

Well if the string is "Hello", then it becomes "ello". So if they were to strncpy 2 characters "Hi" into the buffer, it becomes "Hillo".

– ritlew
Nov 20 '18 at 18:42

AAAAYYYAAA I FOUND OUT THE PROBLEM

– masterCUCK9
Nov 20 '18 at 18:48

@ritlew: And this is one of the reasons I never use strncpy.

– John Bode
Nov 20 '18 at 20:11

add a comment |

In C you have to create a temporary buffer all the time. The most typical way to do this is to write a line of code like this:

char temp_buffer[256];

This buffer will have 256 characters of space available to be able to put strings in.

edited Nov 20 '18 at 18:50

answered Nov 20 '18 at 18:22

ritlew

853411

Comments are not for extended discussion; this conversation has been moved to chat.

– Samuel Liew♦
Nov 21 '18 at 2:15

add a comment |

In C you have to create a temporary buffer all the time. The most typical way to do this is to write a line of code like this:

char temp_buffer[256];

This buffer will have 256 characters of space available to be able to put strings in.

edited Nov 20 '18 at 18:50

answered Nov 20 '18 at 18:22

ritlew

853411

Comments are not for extended discussion; this conversation has been moved to chat.

– Samuel Liew♦
Nov 21 '18 at 2:15

add a comment |

In C you have to create a temporary buffer all the time. The most typical way to do this is to write a line of code like this:

char temp_buffer[256];

This buffer will have 256 characters of space available to be able to put strings in.

edited Nov 20 '18 at 18:50

answered Nov 20 '18 at 18:22

ritlew

853411

In C you have to create a temporary buffer all the time. The most typical way to do this is to write a line of code like this:

char temp_buffer[256];

This buffer will have 256 characters of space available to be able to put strings in.

edited Nov 20 '18 at 18:50

answered Nov 20 '18 at 18:22

ritlew

853411

edited Nov 20 '18 at 18:50

answered Nov 20 '18 at 18:22

ritlew

853411

answered Nov 20 '18 at 18:22

ritlew

853411

answered Nov 20 '18 at 18:22

ritlew

853411

Comments are not for extended discussion; this conversation has been moved to chat.

– Samuel Liew♦
Nov 21 '18 at 2:15

add a comment |

Comments are not for extended discussion; this conversation has been moved to chat.

– Samuel Liew♦
Nov 21 '18 at 2:15

Comments are not for extended discussion; this conversation has been moved to chat.

– Samuel Liew♦
Nov 21 '18 at 2:15

add a comment |

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