Whats the proper way to handle strcpy's return type
0
I created a program which at some point turned to be a buggy one and I can not find a good way to handle situations like in the following program:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int get_user_name( char *const dest, char *const src );
char *foo( char *variable ); /// foo() returns NULL
int main( void )
{
char arr[256] = { 0 };
char buffer[ 256 ] = { 0 };
char *const ptr = foo( arr ); /// foo() returned NULL here
if ( get_user_name( buffer, ptr ) == 0 ) /// get_user_name, should here return 0
{
printf("NULL returned from get_user_name()n" );
exit( EXIT_FAILURE );
}else
{
printf( "Everithing is OK" );
}
}
int get_user_name( char *const dest, char *const src )
{
char *ret = strcpy( dest, src ); /// Here Segfault happens
/// There is no return here because of above Segfault
if ( ret == NULL )
{
return 0;
}else
{
return 1;
}
}
char *foo( char *variable )
{
if ( strlen( variable) < 1 )
{
return NULL; /// Here will return NULL because there is no Length
}
/// Some code here ...
return variable;
}
Here is use a demo function which returns NULL to explain my problem.
I was thinking to check src before past it to strcpy but I really cannot understand why strcpy returns dest without check if src is NULL or not.
Why does this happens because I see that strcpy manual say only about the return of dest and nothing if fails:
RETURN VALUE
The strcpy() and strncpy() functions return a pointer to the destination string dest.
c strcpy
add a comment |
0
I created a program which at some point turned to be a buggy one and I can not find a good way to handle situations like in the following program:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int get_user_name( char *const dest, char *const src );
char *foo( char *variable ); /// foo() returns NULL
int main( void )
{
char arr[256] = { 0 };
char buffer[ 256 ] = { 0 };
char *const ptr = foo( arr ); /// foo() returned NULL here
if ( get_user_name( buffer, ptr ) == 0 ) /// get_user_name, should here return 0
{
printf("NULL returned from get_user_name()n" );
exit( EXIT_FAILURE );
}else
{
printf( "Everithing is OK" );
}
}
int get_user_name( char *const dest, char *const src )
{
char *ret = strcpy( dest, src ); /// Here Segfault happens
/// There is no return here because of above Segfault
if ( ret == NULL )
{
return 0;
}else
{
return 1;
}
}
char *foo( char *variable )
{
if ( strlen( variable) < 1 )
{
return NULL; /// Here will return NULL because there is no Length
}
/// Some code here ...
return variable;
}
Here is use a demo function which returns NULL to explain my problem.
I was thinking to check src before past it to strcpy but I really cannot understand why strcpy returns dest without check if src is NULL or not.
Why does this happens because I see that strcpy manual say only about the return of dest and nothing if fails:
RETURN VALUE
The strcpy() and strncpy() functions return a pointer to the destination string dest.
c strcpy
but I really cannot understand why strcpy returns dest without check if src is NULL or not. - because it says in the standard that it expects a c-string, not aNULL. And I betstrcpy()most times won't return anything but crash almost instantly.
– Swordfish
Nov 24 '18 at 17:34
strcpycauses the abort of your program because one of the parameters is null. Check before calling.
– Paul Ogilvie
Nov 24 '18 at 17:34
@Swordfish I all ready pointed to segfault in the Question. And I also know what the standard say. Expecting something which is a string and relying on that that there will be no NULL without checking, from my side it does not make it the best function. I do know that ascprogrammer I need to check everything againstNULLbut this is not my Question.
– user10533142
Nov 24 '18 at 17:36
add a comment |
0
0
0
I created a program which at some point turned to be a buggy one and I can not find a good way to handle situations like in the following program:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int get_user_name( char *const dest, char *const src );
char *foo( char *variable ); /// foo() returns NULL
int main( void )
{
char arr[256] = { 0 };
char buffer[ 256 ] = { 0 };
char *const ptr = foo( arr ); /// foo() returned NULL here
if ( get_user_name( buffer, ptr ) == 0 ) /// get_user_name, should here return 0
{
printf("NULL returned from get_user_name()n" );
exit( EXIT_FAILURE );
}else
{
printf( "Everithing is OK" );
}
}
int get_user_name( char *const dest, char *const src )
{
char *ret = strcpy( dest, src ); /// Here Segfault happens
/// There is no return here because of above Segfault
if ( ret == NULL )
{
return 0;
}else
{
return 1;
}
}
char *foo( char *variable )
{
if ( strlen( variable) < 1 )
{
return NULL; /// Here will return NULL because there is no Length
}
/// Some code here ...
return variable;
}
Here is use a demo function which returns NULL to explain my problem.
I was thinking to check src before past it to strcpy but I really cannot understand why strcpy returns dest without check if src is NULL or not.
Why does this happens because I see that strcpy manual say only about the return of dest and nothing if fails:
RETURN VALUE
The strcpy() and strncpy() functions return a pointer to the destination string dest.
c strcpy
I created a program which at some point turned to be a buggy one and I can not find a good way to handle situations like in the following program:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int get_user_name( char *const dest, char *const src );
char *foo( char *variable ); /// foo() returns NULL
int main( void )
{
char arr[256] = { 0 };
char buffer[ 256 ] = { 0 };
char *const ptr = foo( arr ); /// foo() returned NULL here
if ( get_user_name( buffer, ptr ) == 0 ) /// get_user_name, should here return 0
{
printf("NULL returned from get_user_name()n" );
exit( EXIT_FAILURE );
}else
{
printf( "Everithing is OK" );
}
}
int get_user_name( char *const dest, char *const src )
{
char *ret = strcpy( dest, src ); /// Here Segfault happens
/// There is no return here because of above Segfault
if ( ret == NULL )
{
return 0;
}else
{
return 1;
}
}
char *foo( char *variable )
{
if ( strlen( variable) < 1 )
{
return NULL; /// Here will return NULL because there is no Length
}
/// Some code here ...
return variable;
}
Here is use a demo function which returns NULL to explain my problem.
I was thinking to check src before past it to strcpy but I really cannot understand why strcpy returns dest without check if src is NULL or not.
Why does this happens because I see that strcpy manual say only about the return of dest and nothing if fails:
RETURN VALUE
The strcpy() and strncpy() functions return a pointer to the destination string dest.
c strcpy
c strcpy
edited Nov 24 '18 at 17:50
asked Nov 24 '18 at 17:29
user10533142
but I really cannot understand why strcpy returns dest without check if src is NULL or not. - because it says in the standard that it expects a c-string, not aNULL. And I betstrcpy()most times won't return anything but crash almost instantly.
– Swordfish
Nov 24 '18 at 17:34
strcpycauses the abort of your program because one of the parameters is null. Check before calling.
– Paul Ogilvie
Nov 24 '18 at 17:34
@Swordfish I all ready pointed to segfault in the Question. And I also know what the standard say. Expecting something which is a string and relying on that that there will be no NULL without checking, from my side it does not make it the best function. I do know that ascprogrammer I need to check everything againstNULLbut this is not my Question.
– user10533142
Nov 24 '18 at 17:36
add a comment |
but I really cannot understand why strcpy returns dest without check if src is NULL or not. - because it says in the standard that it expects a c-string, not aNULL. And I betstrcpy()most times won't return anything but crash almost instantly.
– Swordfish
Nov 24 '18 at 17:34
strcpycauses the abort of your program because one of the parameters is null. Check before calling.
– Paul Ogilvie
Nov 24 '18 at 17:34
@Swordfish I all ready pointed to segfault in the Question. And I also know what the standard say. Expecting something which is a string and relying on that that there will be no NULL without checking, from my side it does not make it the best function. I do know that ascprogrammer I need to check everything againstNULLbut this is not my Question.
– user10533142
Nov 24 '18 at 17:36
but I really cannot understand why strcpy returns dest without check if src is NULL or not. - because it says in the standard that it expects a c-string, not a
NULL. And I bet strcpy() most times won't return anything but crash almost instantly.– Swordfish
Nov 24 '18 at 17:34
but I really cannot understand why strcpy returns dest without check if src is NULL or not. - because it says in the standard that it expects a c-string, not a
NULL. And I bet strcpy() most times won't return anything but crash almost instantly.– Swordfish
Nov 24 '18 at 17:34
strcpy causes the abort of your program because one of the parameters is null. Check before calling.– Paul Ogilvie
Nov 24 '18 at 17:34
strcpy causes the abort of your program because one of the parameters is null. Check before calling.– Paul Ogilvie
Nov 24 '18 at 17:34
@Swordfish I all ready pointed to segfault in the Question. And I also know what the standard say. Expecting something which is a string and relying on that that there will be no NULL without checking, from my side it does not make it the best function. I do know that as
c programmer I need to check everything against NULL but this is not my Question.– user10533142
Nov 24 '18 at 17:36
@Swordfish I all ready pointed to segfault in the Question. And I also know what the standard say. Expecting something which is a string and relying on that that there will be no NULL without checking, from my side it does not make it the best function. I do know that as
c programmer I need to check everything against NULL but this is not my Question.– user10533142
Nov 24 '18 at 17:36
add a comment |
1 Answer
1
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1
You handle the situation by first checking the parameters are not null. For example:
int get_user_name( char *dest, char *const src )
{
char *ret;
if (dest==0 || src==0) return 0;
strcpy( dest, src );
return 1;
}
Note also that dest cannot be const because it is written to.
answered Nov 24 '18 at 17:39
Paul OgilviePaul Ogilvie
18.3k21235
Well, that means thatstrcpyworks after I know 100% thatdestandsrcare OK. Thank you
– user10533142
Nov 24 '18 at 17:40
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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oldest
votes
1
You handle the situation by first checking the parameters are not null. For example:
int get_user_name( char *dest, char *const src )
{
char *ret;
if (dest==0 || src==0) return 0;
strcpy( dest, src );
return 1;
}
Note also that dest cannot be const because it is written to.
answered Nov 24 '18 at 17:39
Paul OgilviePaul Ogilvie
18.3k21235
Well, that means thatstrcpyworks after I know 100% thatdestandsrcare OK. Thank you
– user10533142
Nov 24 '18 at 17:40
add a comment |
1
You handle the situation by first checking the parameters are not null. For example:
int get_user_name( char *dest, char *const src )
{
char *ret;
if (dest==0 || src==0) return 0;
strcpy( dest, src );
return 1;
}
Note also that dest cannot be const because it is written to.
answered Nov 24 '18 at 17:39
Paul OgilviePaul Ogilvie
18.3k21235
Well, that means thatstrcpyworks after I know 100% thatdestandsrcare OK. Thank you
– user10533142
Nov 24 '18 at 17:40
add a comment |
1
1
1
You handle the situation by first checking the parameters are not null. For example:
int get_user_name( char *dest, char *const src )
{
char *ret;
if (dest==0 || src==0) return 0;
strcpy( dest, src );
return 1;
}
Note also that dest cannot be const because it is written to.
answered Nov 24 '18 at 17:39
Paul OgilviePaul Ogilvie
18.3k21235
You handle the situation by first checking the parameters are not null. For example:
int get_user_name( char *dest, char *const src )
{
char *ret;
if (dest==0 || src==0) return 0;
strcpy( dest, src );
return 1;
}
Note also that dest cannot be const because it is written to.
answered Nov 24 '18 at 17:39
Paul OgilviePaul Ogilvie
18.3k21235
answered Nov 24 '18 at 17:39
Paul OgilviePaul Ogilvie
18.3k21235
answered Nov 24 '18 at 17:39
Paul OgilviePaul Ogilvie
18.3k21235
answered Nov 24 '18 at 17:39
Paul OgilviePaul Ogilvie
18.3k21235
18.3k21235
Well, that means thatstrcpyworks after I know 100% thatdestandsrcare OK. Thank you
– user10533142
Nov 24 '18 at 17:40
add a comment |
Well, that means thatstrcpyworks after I know 100% thatdestandsrcare OK. Thank you
– user10533142
Nov 24 '18 at 17:40
Well, that means that
strcpy works after I know 100% that dest and src are OK. Thank you– user10533142
Nov 24 '18 at 17:40
Well, that means that
strcpy works after I know 100% that dest and src are OK. Thank you– user10533142
Nov 24 '18 at 17:40
add a comment |
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but I really cannot understand why strcpy returns dest without check if src is NULL or not. - because it says in the standard that it expects a c-string, not a
NULL. And I betstrcpy()most times won't return anything but crash almost instantly.– Swordfish
Nov 24 '18 at 17:34
strcpycauses the abort of your program because one of the parameters is null. Check before calling.– Paul Ogilvie
Nov 24 '18 at 17:34
@Swordfish I all ready pointed to segfault in the Question. And I also know what the standard say. Expecting something which is a string and relying on that that there will be no NULL without checking, from my side it does not make it the best function. I do know that as
cprogrammer I need to check everything againstNULLbut this is not my Question.– user10533142
Nov 24 '18 at 17:36