Allocating memory for a pointer passed from another function

[Adam Baker]

I'm sure this is a fairly basic question. I want a pointer in main()
to be passed to a function, and to allocate memory for the pointer in
the function, and then have it available to main after.

This does not work:

#include <stdio.h>

void func(float *a)
{
a = malloc(5*sizeof(float));
*(a+5) = 15.0f;
}

int main()
{
float *ptr;
func(ptr);
printf("%g\n",*(ptr+5)); // prints something other than 15
return 0;
}

This sort of makes sense, since I'm passing ptr as an argument. Is
there a way to pass ptr so that it behaves as expected? Or do I have
to do this?

#include <stdio.h>

float* func()
{
float *a = malloc(5*sizeof(float));
*(a+5) = 15.0f;
return a;
}

int main()
{
float *ptr;
ptr = func();
printf("%g\n",*(ptr+5));
return 0;
}
Thanks,
Adam

[Francine.Neary]

On Jan 10, 9:10 pm, Adam Baker <[e-mail]:

I'm sure this is a fairly basic question. I want a pointer in main()
to be passed to a function, and to allocate memory for the pointer in
the function, and then have it available to main after.

This does not work:

#include <stdio.h>

void func(float *a)
{
a = malloc(5*sizeof(float));
*(a+5) = 15.0f;

}

int main()
{
float *ptr;
func(ptr);
printf("%g\n",*(ptr+5)); // prints something other than 15
return 0;

}

This sort of makes sense, since I'm passing ptr as an argument. Is
there a way to pass ptr so that it behaves as expected?

Function arguments are always passed by value in C.

You can pass a pointer to a pointer instead:

#include <stdio.h>
#include <stdlib.h>

void func(float **a)
{
*a = malloc(6 * sizeof **a);
if(*a)
*(*a+5) = 15.0f;
else {
fputs("No memory\n", stderr);
exit(EXIT_FAILURE);
}
}

int main(void)
{
float *ptr;
func(&ptr);
printf("%g\n", *(ptr+5));
free(ptr);
return 0;
}

(Note some other subtle difference from your code, too.)

[Default User]

Adam Baker wrote:

I'm sure this is a fairly basic question. I want a pointer in main()
to be passed to a function, and to allocate memory for the pointer in
the function, and then have it available to main after.

This does not work:

#include <stdio.h>

void func(float *a)
{
a = malloc(5*sizeof(float));
*(a+5) = 15.0f;
}

int main()
{
float *ptr;
func(ptr);
printf("%g\n",*(ptr+5)); // prints something other than 15
return 0;
}

This is essentially FAQ 4.8:

<http://c-faq.com/ptrs/passptrinit.html>


Brian

--
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newsgroup, or:
<http://www.caliburn.nl/topposting.html>

[CBFalconer]

Adam Baker wrote:

>
I'm sure this is a fairly basic question. I want a pointer in main()
to be passed to a function, and to allocate memory for the pointer in
the function, and then have it available to main after.

This does not work:

#include <stdio.h>

void func(float *a) {
a = malloc(5*sizeof(float));
*(a+5) = 15.0f;
}

int main() {
float *ptr;
func(ptr);
printf("%g\n",*(ptr+5)); // prints something other than 15
return 0;
}

This sort of makes sense, since I'm passing ptr as an argument. Is
there a way to pass ptr so that it behaves as expected? Or do I have
to do this?

Of course not. float receives a, of type float*, by value. Any
changes are not visible in main. You want:

void func(float* *a) {
if (*a = malloc(5 * sizeof(float))
*(*a + 4) = 15.0f;
/* ^-- *a only holds indices 0 thru 4 */
}

and in main you call "func(&ptr);". Better would be to have func
return the value of a, eliminating all the double **, and making
the call "ptr = func(ptr);".

--
Chuck F (cbfalconer at maineline dot net)
<http://cbfalconer.home.att.net>
Try the download section.

--
Posted via a free Usenet account from http://www.teranews.com

[henry]

On Jan 10, 1:27 pm, [e-mail] wrote:

On Jan 10, 9:10 pm, Adam Baker <[e-mail]:

I'm sure this is a fairly basic question. I want apointerin main()
to be passed to afunction, and to allocate memory for thepointerin
thefunction, and then have it available to main after.

This does not work:

#include <stdio.h>

void func(float *a)
{
a = malloc(5*sizeof(float));
*(a+5) = 15.0f;

}

int main()
{
float *ptr;
func(ptr);
printf("%g\n",*(ptr+5)); // prints something other than 15
return 0;

}

This sort of makes sense, since I'm passing ptr as anargument. Is
there a way to pass ptr so that it behaves as expected?

Functionarguments are always passed by value in C.

You can pass apointerto apointerinstead:

#include <stdio.h>
#include <stdlib.h>

void func(float **a)
{
*a = malloc(6 * sizeof **a);
if(*a)
*(*a+5) = 15.0f;
else {
fputs("No memory\n", stderr);
exit(EXIT_FAILURE);
}

}

int main(void)
{
float *ptr;
func(&ptr);
printf("%g\n", *(ptr+5));
free(ptr);
return 0;

}

(Note some other subtle difference from your code, too.)

Another example is strcpy(char *p, const char *s); why
it is not strcpy(char **p,const char *s)?

[Keith Thompson]

henry <he**********@yahoo.comwrites:

On Jan 10, 1:27 pm, [e-mail] wrote:

>On Jan 10, 9:10 pm, Adam Baker <[e-mail]:

[...]
[attribution lost]

>Function arguments are always passed by value in C.

You can pass a pointer to a pointer instead:

[...]

>void func(float **a)
{
*a = malloc(6 * sizeof **a);
if(*a)
*(*a+5) = 15.0f;
else {
fputs("No memory\n", stderr);
exit(EXIT_FAILURE);
}

[...]

>
Another example is strcpy(char *p, const char *s); why
it is not strcpy(char **p,const char *s)?

Because strcpy doesn't need to modify the pointer to the target
string; it only modifies the content of the string itself.

--
Keith Thompson (The_Other_Keith) <ks***@mib.org>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"