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Guest
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 Posted: Sat May 12, 2007 9:11 am Post subject: Array of Pointers and Freeing problem |
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hi,
Following is problematic Code
main()
{
char *ptr[1]={NULL};
ftn(ptr);
}
void ftn(**data)
{
*data=malloc(4);
memset(*data,0,4);
strcpy(*data,"HAI");
} |
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Richard Heathfield
Guest
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| Posted: Sat May 12, 2007 9:11 am Post subject: Re: Array of Pointers and Freeing problem |
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anand1603 (AT) yahoo (DOT) com said:
| Quote: | hi,
Following is problematic Code
main()
{
char *ptr[1]={NULL};
ftn(ptr);
}
void ftn(**data)
{
*data=malloc(4);
memset(*data,0,4);
strcpy(*data,"HAI");
}
|
This won't compile either.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www. |
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Ian Collins
Guest
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| Posted: Sat May 12, 2007 9:11 am Post subject: Re: Array of Pointers and Freeing problem |
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anand1603 (AT) yahoo (DOT) com wrote:
| Quote: | hi,
Following is problematic Code
Indeed it is, it won't compile. |
| Quote: | main()
{
char *ptr[1]={NULL};
ftn(ptr);
}
void ftn(**data)
{
*data=malloc(4);
memset(*data,0,4);
strcpy(*data,"HAI");
}
|
--
Ian Collins. |
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Martin Ambuhl
Guest
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| Posted: Sat May 12, 2007 9:11 am Post subject: Re: Array of Pointers and Freeing problem |
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anand1603 (AT) yahoo (DOT) com wrote:
| Quote: | hi,
Following is problematic Code
|
Yes, it is.
Start by paying attention to the following changes. Then decide what I
have left undone.
/* You need these headers */
#include <stdlib.h>
#include <string.h>
/* Without the following, ftn (in main) is explicitly declared to
return an int, which conflicts with its later redefinition */
void ftn(char *data[]);
/* main returns an int. You should say so. Not doing so is an error
for the C standard from 1999 on, and a bad idea for the earlier
standard. */
int main(void)
{
char *ptr[1] = { NULL };
ftn(ptr);
/* You allocated space for ptr[0] to point at. free it */
free(*ptr);
/* main returns an int. You should do so. Not doing so is an error
for the C standard up to 1999, and a bad idea for the later
standard. */
return 0;
}
/* You have specified no type for the argument to ftn, do so: */
void ftn(char *data[])
{
*data = malloc(4);
/* you should always check that malloc succeeds */
if (!*data)
return;
#if 0
/* The following memset is completely worthless. It has been
commented out. */
memset(*data, 0, 4);
#endif
strcpy(*data, "HAI");
} |
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Barry Schwarz
Guest
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| Posted: Sun May 13, 2007 5:03 am Post subject: Re: Array of Pointers and Freeing problem |
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On Sat, 12 May 2007 18:07:04 -0400, CBFalconer <cbfalconer (AT) yahoo (DOT) com>
wrote:
| Quote: | Richard Heathfield wrote:
anand1603 (AT) yahoo (DOT) com said:
Following is problematic Code
main()
{
char *ptr[1]={NULL};
ftn(ptr);
}
void ftn(**data)
{
*data=malloc(4);
memset(*data,0,4);
strcpy(*data,"HAI");
}
This won't compile either.
However, this will (untested). (To OP) Study the difference.
#include <stdlib.h
#include <string.h
void ftn(**data) {
|
Doesn't the argument need a type? I think you meant
void ftn(char **data){
| Quote: | *data = malloc(4);
memset(*data, 0, 4);
strcpy(*data, "HAI");
}
int main(void) {
char *ptr[1] = {NULL};
ftn(ptr);
return 0;
}
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