Hexadecimal array in pointer

Question

 # include <stdio.h>
 int main()
 {
    char p[]={0x01,0x02,0x03,0x04};
    int *q = p;
    printf("%x",*q);
    return 0;

  }

When I run the above code, the answer is 4030201. Why?

"Why?" Because you got unlucky. Had the code printed 'UB detected. Please ask on SO why' you had got lucky! — pmg
– pmg, Commented Oct 6, 2019 at 13:58
int *q = p; is illegal in C. Your compiler told you about it. The rest is just experiments with undefined code. — AnT stands with Russia
– AnT stands with Russia, Commented Oct 6, 2019 at 14:01

Alexander Ushakov · Accepted Answer · 2019-10-06 14:07:38Z

1

It is because you compile and run your code on little-endian architecture machine (x86 for example)

At first you put 4 bytes in memory in order: 01 02 03 04. Than convert pointer to this array to pointer to int. On little-endian machine memory block 01 02 03 04 represents integer value 0x04030201 which is printed on next step.

See https://en.m.wikipedia.org/wiki/Endianness for more information

answered Oct 6, 2019 at 14:07

Alexander Ushakov

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