C Exam Char Array

Question

I am doing previous year C programming exam. And I came up with this:

A program (see below) defines the two variables x and y.

It produces the given output. Explain why the character ‘A’ appears in the output of variable x.

Program:

#include <stdio.h>
main ()
{
    char x[6] = "12345\0";
    char y[6] = "67890\0";
    y[7]='A';
    printf("X: %s\n",x);
    printf("Y: %s\n",y);
}   

Program output: X: 1A345 Y: 67890

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It has pretty high points (7). And I don't know how to explain it in detail. My answer would be:

char array (y) only have 6 chars allocated so changing 7th character will change whatever is after that in stack.

Any help would highly appreciated! (I'm only 1st year)

Answer 1

Your formal answer should be that this program yields undefined behavior.

The C-language standard does not define the result of an out-of-bound access operation.

With char y[6], by reading from or writing into y[7], this is exactly what you are doing.

Some compilers may choose to allocate array x[6] immediate after array y[6] in the stack.

So by writing 'A' into y[7], this program might indeed write 'A' into x[1].

But the standard does not dictate that, so it depends on compiler implementation.

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As others have implied on previous comments to your question, if it was really given on a formal exam, then you may want to consider continuing your studies elsewhere...

Answer 2

The classic stack corruption problem in C. With the help of a debugger, you will find that your frame stack will look like this after the original assignments:

67890\012345\0

y points to the char 6. y[7] means 7 positions after that (2). So y[7] = 'A' replaces the char 2.

Access array beyond bound is undefined in the C standard, just one more quirk of C to be aware of. Some references:

• Understanding stack corruption
• Why do compilers not warn about out-of-bounds static array indices?