Martin Thoma

What is the output of the following programm?

#include<stdio.h>

int main() {
	printf("%i\n", -13>>1);
}

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Short answer

-7

Long answer

Signed integers are two’s complement binary values that can be used to represent both positive and negative
integer values.

Source: Intels IA-32 Architectures Software Developer’s Manuals, page 83

When you want to get the two’s complement representation of -13, you have to get the binary representation of 13, invert the digits and add one. As 13 is 1101 in binary, -13 looks like this on a 32 bit machine:
1111.1111.1111.1111.1111.1111.1111.0011

So two results might be logical:
0111.1111.1111.1111.1111.1111.1111.1001

or

1111.1111.1111.1111.1111.1111.1111.1001

Lets get the assembly code:

gcc -S -O0 test.c

	.file	"test.c"
	.section	.rodata
.LC0:
	.string	"%i\n"
	.text
	.globl	main
	.type	main, @function
main:
.LFB0:
	.cfi_startproc
	pushl	%ebp
	.cfi_def_cfa_offset 8
	.cfi_offset 5, -8
	movl	%esp, %ebp
	.cfi_def_cfa_register 5
	andl	$-16, %esp
	subl	$16, %esp
	movl	$-7, 4(%esp)
	movl	$.LC0, (%esp)
	call	printf
	leave
	.cfi_restore 5
	.cfi_def_cfa 4, 4
	ret
	.cfi_endproc
.LFE0:
	.size	main, .-main
	.ident	"GCC: (Ubuntu/Linaro 4.7.2-2ubuntu1) 4.7.2"
	.section	.note.GNU-stack,"",@progbits

In line 18 you can see that the bitshift already happened. When you introduce a variable for a, you can see that the compiler makes use of the assembly command sarl %eax or sarl $2, %eax when you shift by two.

When you take a look at Oracles IA-32 Assembly Language Reference Manual, page 56, you find:

sar right shifts (signed divides) a byte, word, or long value for a count specified by
an immediate value and stores the quotient in that byte, word, or long respectively.
The second variation right shifts by a count value specified in the CL register. sar
rounds toward negative infinity; the high-order bit remains unchanged.

This means, 1111.1111.1111.1111.1111.1111.1111.1001 is correct. And that’s -7.