Using char pointer to print value of an integer variable

Question

So am trying to write a trivial program just for fun. Intent is to print the value of an integer using char pointer

So you can see I have an integer a, which I print and then I use a char* b to get the higher byte location of the integer and then get the lower byte using post-increment operation of char* b

But on https://code.sololearn.com the line which prints *b and *b++, throws warning for *b++. Am interested in knowing what is the way to remove that warning.

#include <stdio.h>

int main(void) {

    int a = 0xaabb;

    printf("%x\n",a);

    unsigned char *b = (unsigned char*) &a;

    printf("%x%x\n",*b, *b++);

    return 0;
}

Use a union, e.g. typedef union { unsigned char bytes[4]; int n; } uc2n; Then uc2n convert = { .n = 0xaabb }; now you can use convert.bytes[0] - convert.bytes[3] to access each byte of the int. — David C. Rankin
– David C. Rankin, Commented Feb 28, 2020 at 19:58
You are changing the approach completely. I understand the array approach. I was trying to understand warning with post-increment in the pointer approach — Tahseen
– Tahseen, Commented Feb 28, 2020 at 20:07
It would be nice to know what the exact error message is. Beyond that, since you've done a post-increment, your final print statement will print *b twice. I am not sure if that was your intent or not. — Frank Merrow
– Frank Merrow, Commented Feb 28, 2020 at 20:11
@FrankMerrow not necessarily. Not only is the order of evaluation not specified, but the post-increment might be done before the other b is placed as an argument. A post-increment happens at some time between that b being used, and the next sequence point. — Weather Vane
– Weather Vane, Commented Feb 28, 2020 at 20:13

David C. Rankin · Accepted Answer · 2020-02-28 20:18:03Z

2

The problem you are having is due to the order of evaluation of the parameters passed to printf being unspecified. You are attempting to pass *b, *b++. Since the order in which the parameters are evaluated is not specified, the compiler cannot guarantee what the value of *b will be. If *b++ is evaluated first, then the first parameter will be one greater than the second. If *b++ is evaluated second, the both parameters are the same.

That is the reason you receive a warning similar to:

warning: operation on ‘b’ may be undefined [-Wsequence-point]

The sequence of evaluation of the parameters is unspecified making the operation on b undefined.

If you want to correct the warning without using a union, then you can simply add one to b, e.g.

#include <stdio.h>

int main(void) {

    int a = 0xaabb;

    printf("%x\n",a);

    unsigned char *b = (unsigned char*)&a;

    printf ("%x%x\n",*b, *(b + 1));

}

or use a second pointer, e.g.

#include <stdio.h>

int main(void) {

    int a = 0xaabb;

    printf("%x\n",a);

    unsigned char *b = (unsigned char*)&a;
    unsigned char *c = b + 1;

    printf ("%x%x\n",*b, *c);

}

edited Feb 28, 2020 at 20:18

answered Feb 28, 2020 at 20:09

David C. Rankin

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5 Comments

Weather Vane Over a year ago

If the second argument is evaluated first, the post-increment is not guaranteed to happen before the other b is accessed because there is no sequence point yet.

Tahseen Over a year ago

Can you elaborate "Since the order in which the parameters are evaluated is not specified" ? printf knows first *b would come and then *b++

Andrew Henle Over a year ago

@Tahseen printf knows first *b would come and then *b++ No, it doesn't "know" that. Just because *b is to the left of *b++ in the code you wrote does not mean that *b will be evaluated before *b++. See stackoverflow.com/questions/4176328/…

David C. Rankin Over a year ago

You have printf ("..", param1, param2). The order in which param1 and param2 are evaluated is not specified by the standard. So either can happen. If the order of evaluation is param1, param2, then the post increment is taken after the parameter has been evaluated and param1 == param2 for the call, If the order is param2, param1 the post increment is applied and the parameters seen by printf differ -- but you have no guarantee of knowing which will occur.

Weather Vane Over a year ago

@Tahseen printf neither knows nor cares which argument was evaluated or placed first. It only knows where to find them.

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