CODEWITHSUNDEEP
c++eclipse
Sreram
Sreram

Reputation: 501

Why is this program's output unexpected?

The following code when executed in eclipse, running on Ubuntu and compiled using g++ compiler, provides unexpected results.

code:

#include <iostream>
int main()
{
    unsigned int a=5555;
    std::cout << (unsigned int)(((char*)&a)[0]) <<  "\n";
    std::cout << (unsigned int)(((char*)&a)[1]) <<  "\n";
    std::cout << (unsigned int)(((char*)&a)[2]) <<  "\n";
    std::cout << (unsigned int)(((char*)&a)[3]) <<  "\n";

  return 0;
}

I am trying to treat the variable a as an array of integers each of one byte size. When I execute the program, this is what I get as output:

output:

4294967219
21
0
0

question:

Why is the first value displayed so large (here int is of size 32 bits or 4 bytes). So each of the output values should obviously be no greater than 255 right? And why are the last three values zero? Or why I am getting the wrong result?

I also got the same result when tested in code::blocks, running the same compiler.

Upvotes: 1

Views: 67

Answers (2)

Dolda2000
Dolda2000

Reputation: 25855

This is because of sign-extension. Let's look at your unsigned int a in memory:

b3 15 00 00

When you cast the first byte from a signed char to an unsigned int, the cast from char to int happens before the conversion from signed to unsigned, and therefore, the sign bit is extended, and the result (0xffffffb3) is what you see on your first line.

Try casting to an unsigned char * instead of a char *.

Upvotes: 5

Sam Varshavchik
Sam Varshavchik

Reputation: 118425

This because char is a signed integer type.

Decimal 5555 is hexadecimal 0x15b3.

The 0xb3 byte, when sign-extended to a signed int, becomes 0xffffffb3.

0xffffffb3 interpreted as an unsigned int is 4294967219 in decimal.

Upvotes: 6

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