Can't modify values at addresses allocated by malloc()

Question

In the following code:

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

int main(int argc, char *argv[])
{

    float *a; 
    printf("%x\n",&a);
    a = malloc(100*sizeof(float));
    printf("%x\n",a);
    *a=5;
    printf("%x\n",*a);
}

I expect to see "5" in the final output, but all I see is zero. How can I modify those memory space?

Why are you using an unsigned hexadecimal integer specifier to print a `float`? — David Ranieri, Apr 20 '16 at 08:33
oh god :) for some reason I thought I could print everything with %x — C K, Apr 20 '16 at 08:34
... and pointer values should be printed with `%p` and not `%x`. Read about `printf`. — Jabberwocky, Apr 20 '16 at 08:36
@Boiethios: "*%x is a pointer*" no, the `x` conversion specifier is for unsigned integers. — alk, Apr 20 '16 at 08:52

score 4 · Accepted Answer · edited May 23 '17 at 11:59

I expect to see "5" in the final output, but all I see is zero. How can I modify those memory space?

Use "%f" (or "%g" etc appropriate identifier) to print the value. (Or use explicit casting to unsigned int and print sizeof(double) * CHAR_BIT bits)

Printing a double argument with "%u" printf format specifier leads to undefined behavior. Related post: Understanding implicit conversions for printf

Can't modify values at addresses allocated by malloc()

You should be able to modify the contents on conforming implementation if a is not NULL. There is problem in presenting the data.

user3078414 · Answer 2 · 2016-04-20T10:03:47.197

1

I believe this is the working answer to your question. It seems to be about printf() format specifiers and entering float value as float, not as integer:

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

int main(int argc, char *argv[])
{
float *a; 
printf("%p\n",(void*)a);
a = malloc(100*sizeof(float));
printf("%p\n",(void*)a);
if(a != NULL) *a=5.0;
printf("%f\n",*a);
}

The terminal output then looks like this:

> ./a.out
0x0
0x1031009a0
5.000000
>

edited Apr 20 '16 at 10:03

answered Apr 20 '16 at 09:02

user3078414

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1

Don't print uninitialized pointer. Cast pointer to `void *` before printing. – Mohit Jain Apr 20 '16 at 09:20
1

Thanks for this suggestion @Mohit Jain - I improved my reply by including it. Although "printfing" uninitialized pointers provenly works with my BSD UNIX, one always has to be extra cautious with _uninitialized pointers_ in portable code. (-: – user3078414 Apr 20 '16 at 09:27

Can't modify values at addresses allocated by malloc()

2 Answers2