Unable to understand output produced after typecasting in c in little endian machine

Question

Below is the C program and the output is 201. I have no idea how this is possible. Please explain this.

#include <stdio.h>
  int main() {
    int number = 288;
    char * ptr;
    ptr = (char * ) & number;
    printf("%x", * ptr);
    ptr++;
    printf("%x", * ptr);
    return 0;
  }

This is in case of little endian machine. This question was asked in the MCQ contest on geeksforgeeks. But they provided no explaination of it.

UB . Printf if expeting unsigned int but is getting the char. — 0___________, Jan 26 '19 at 13:27
https://stackoverflow.com/questions/14504148/what-can-happen-if-printf-is-called-with-a-wrong-format-string before asking use search — 0___________, Jan 26 '19 at 13:31
@P__J__ I wasn't aware of exact question to ask that what is this weird behaviour according to me. So I search for my code but couldn't find anything. That is why I asked. — Brij Raj Kishore, Jan 26 '19 at 13:39
@BrijRajKishore There is no undefined behavior in this case. It is a valid use of printf. See my answer. — Bernd Elkemann, Jan 26 '19 at 13:45

score 2 · Accepted Answer · answered Jan 26 '19 at 13:34

2

The 201 is perfectly reasonable (on little-endian systems):

The variable number starts with these two bytes (on little-endian systems): 0x20, 0x01.

answered Jan 26 '19 at 13:34

Bernd Elkemann

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... so, between your two `%x` you should probably insert a comma. And also insert 0x to make it visible that these are bytes printed as hex – Bernd Elkemann Jan 26 '19 at 13:53
1

I totally ovelooked hexa decimal specifier. I was interpreting it as decimal. Thanks a lot. – Brij Raj Kishore Jan 26 '19 at 13:53

Unable to understand output produced after typecasting in c in little endian machine

1 Answers1