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up vote 0 down vote favorite 
#include <stdio.h>
#include <stdlib.h>

int test(void)
{
  int a=0,b=0;
  char buf[4];
  gets(buf);
  printf("a:%d b:%d",a,b);
}

int main()
{
  test();
  return 0;
}

Question is why with input:aaaaa a is becoming 97 instead of b?From the way variables are declared inside test when buf overflows shouldn't it affect first b and then a?

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did you find an answer? you should mark the most helpful as the selected answer if you did. –  cacho Jun 16 '13 at 20:44

2 Answers 2

up vote 1 down vote

a and b variables will not be necessarily contiguous to the variable buf. Consequently, the overflow of the variable buf, has nothing to do with the possible values of a and b. The behaviour will be undefined.

However, it is important to mention that the C standard will store all of the arrays, such as buf, in continuous memory location.

Here you can check the documentation:

An array is a series of elements of the same type placed in contiguous memory locations that can be individually referenced by adding an index to a unique identifier.

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up vote 0 down vote

Undefined behaviour is undefined. There's nothing in the language standard about the relative locations of different variables in a function, and there's definitely no guarantees about what will happen in a buffer overflow situation.

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