Manipulating pointer values inside functions taking pointers as arguments

Question

I have read a lot of questions of stackoverflow, but couldn't find any solutions on how to deal with this problem of allocating and manipulating pointers inside functions: Can anybody please tell me what's wrong with this code? (I want to allocate and assign values to *D through pointerpass and print the same through pointerprint)

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

float *D;

void pointerpass(float **ptr1)
{
    *ptr1=(float*)malloc(3*sizeof(float));
    *(ptr1+0)=1.33;
    *(ptr1+1)=2.33;
    *(ptr1+2)=3.33;
}

void pointerprint(float **ptr2)
{
    int j=0;
    for (j=0;j<3;j++)
        printf("\n%f\n",*(ptr2+j));
}

int main() 
{
    pointerpass(&D);
    pointerprint(&D);
    return 0;
}

And why are you making life so difficult. You can pass back the result of `malloc`] — Ed Heal, Jun 29 '14 at 19:51
In C, `*ptr1=(float*)malloc(3*sizeof(float));` - the `(float*)` cast is redundant. — Johann Gerell, Jun 29 '14 at 19:52
@EdHeal Indent'ed' it. I am new to C. Can you please tell me why should we pass the result of malloc? — G.One, Jun 29 '14 at 20:00
@Deduplicator Why doesn't *(a+b) work, but Bluepixy's code in the answer below works? — G.One, Jun 29 '14 at 20:01
[Don't cast the result of malloc (and friends)](http://stackoverflow.com/questions/605845). Most important, a cast says to the compiler: Shut up and do as I say. Bluepixy gave you corrected code, he didn't point out any errors, syntax- or style-points. — Deduplicator, Jun 29 '14 at 20:31

score 1 · Answer 1 · answered Jun 29 '14 at 20:19

1

Here we go

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

float * pointerpass(){
   float *ret = malloc(3*sizeof(float));
   ret[0] = 1.33f;
   ret[1] = 2.33f;
   ret[2] = 3.33f;
   return ret;
}
void pointerprint(float *array) {
    int j=0;
    for (j=0;j<3;j++) {
        printf("\n%f\n",array[j]);
    }
}

int main() {
   float *x = pointerpass();
   pointerprint(x);
   free(x); // We do not like memory leaks
   return 0;
}

answered Jun 29 '14 at 20:19

Ed Heal

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17
87
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Finally understood it!! Thanks a lot. – G.One Jun 29 '14 at 21:04

score 0 · Accepted Answer · answered Jun 29 '14 at 19:54

0

void pointerpass(float **ptr1){
    *ptr1=(float*)malloc(3*sizeof(float));
    (*ptr1)[0]=1.33;
    (*ptr1)[1]=2.33;
    (*ptr1)[2]=3.33;
}
void pointerprint(float **ptr2){
    int j=0;
    for (j=0;j<3;j++)
        printf("\n%f\n", (*ptr2)[j]);
}

answered Jun 29 '14 at 19:54

BLUEPIXY

39,699
7
33
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Programming language can be represented exactly. – BLUEPIXY Jun 29 '14 at 20:01
@Deduplicator why can't we cast malloc? – G.One Jun 29 '14 at 20:02
@user3748848 , No problem just redundant it simply **correct** case. But cast it wrong often human error is made. Not required even though do it. – BLUEPIXY Jun 29 '14 at 20:07
Well, it is not a code of enough explanation needed exaggerated. ;D – BLUEPIXY Jun 30 '14 at 09:30

Manipulating pointer values inside functions taking pointers as arguments

2 Answers2