output of the Program, in c++,&int to char* casting

Question

Can somebody explain the output of the following program:

int main()
{
    int i=512;
    char *c=(char*)&i;
    c[0]=1;
    cout<<"i is:"<<i<<endl;

    return 0;
}

output is:513

the output is fixed in case the architecture and the compiler is knowm, but i was not able to understand the output... — sandeep sandha, Oct 30 '13 at 11:12

NPE · Accepted Answer · 2013-10-30T12:28:14.520

3

The output of your program unspecified. In practice, it depends on the endianness of your platform and the width of the int type.

Your platform is little-endian. Let's for simplicity assume that int is 32 bits wide.

512₁₀ is 0x00000200 in hex. This is stored in memory as

00 02 00 00

Your code modifies the first byte to 01. This changes the int to 0x00000201, which is 513 decimal.

edited Oct 30 '13 at 12:28

answered Oct 30 '13 at 10:55

NPE

486,780
108
951
1,012

2

This is implementation specific value. Aliasing with `char` is explicit exception to strict aliasing. – Maciej Piechotka Oct 30 '13 at 10:58
Why is it undefined behavior? As long as the endianess is known, I don't see why it should be not well-defined what's happening. Can you explain why it is UB then? – leemes Oct 30 '13 at 10:58
I've reworded the answer. I believe it is correct now. Let me know if you think otherwise. – NPE Oct 30 '13 at 11:02
Well, if it *really* is UB then the previous version was correct. I don't know if it is UB, I only wanted to know why you thought it would be. – leemes Oct 30 '13 at 11:05
3

@leemes *As long as the endianess is known.* Where in the standard is endianness mentioned? – David Heffernan Oct 30 '13 at 11:17

David Heffernan · Answer 2 · 2013-10-30T11:04:39.927

2

The program exhibits unspecified behaviour, dependent on the architecture of the machine. To predict and reason about the output requires knowledge of the compiler and the target architecture.

edited Oct 30 '13 at 11:04

answered Oct 30 '13 at 10:55

David Heffernan

601,492
42
1,072
1,490

1

A reason would be nice. – leemes Oct 30 '13 at 10:56
@leemes A reason for what? A reason as to why this is UB? – David Heffernan Oct 30 '13 at 10:58
2

Exactly. Can you give a quote of some paragraph? Or what *exactly* about this program is UB? – leemes Oct 30 '13 at 10:59
@leemes for a start, what reinterpret_cast (or c-style cast) does is UB – BЈовић Oct 30 '13 at 11:04
@leemes I don't know whether or not the program formally exhibits UB. – David Heffernan Oct 30 '13 at 11:05

Baldrick · Answer 3 · 2013-10-31T07:25:16.997

2

Explanation:

int main()
{
    /* Creates int equal to 512 */
    int i=512;

    /* Creates a char pointer, and points this at i */
    char *c=(char*)&i;

    /* Overwrites the lowest byte of the 4 byte int with 1 */
    /* This sets the lowest bit of the int, which adds 1 */
    c[0]=1;

    /* Displays the updated int */
    cout<<"i is:"<<i<<endl;

    return 0;
}

Exactly which part of the int gets overwritten depends on the endianness of the platform you're compiling for. Given the final result of 513, your system is clearly little-endian.

edited Oct 31 '13 at 07:25

answered Oct 30 '13 at 10:57

Baldrick

11,712
2
31
35

You should probably qualify your explanation as applying specifically to little endian architectures. – Paul R Oct 31 '13 at 07:05

output of the Program, in c++,&int to char* casting

3 Answers3