Line 16: Line 16:
 
int *a;  
 
int *a;  
 
</syntaxhighlight>
 
</syntaxhighlight>
This declares a as an integer pointer, meaning a can point to a memory address which contains an int
+
This declares a as an integer pointer, meaning 'a' can point to any memory address which contains an int
  
 
<syntaxhighlight lang="c">
 
<syntaxhighlight lang="c">

Revision as of 13:41, 8 December 2013

<syntaxhighlight lang="c">

  1. include<stdio.h>

int main() {

int *a;
*a=5;
printf("%d",a);
return 0;

}

</syntaxhighlight>

Solution

<syntaxhighlight lang="c"> int *a; </syntaxhighlight> This declares a as an integer pointer, meaning 'a' can point to any memory address which contains an int

<syntaxhighlight lang="c">

  • a = 5;

</syntaxhighlight> This makes the content of the address pointed to by a 5. But a is not pointing to any valid address (int *a assigns garbage value to a) and hence this assignment can cause segmentation fault



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<syntaxhighlight lang="c">

  1. include<stdio.h>

int main() {

int *a;
*a=5;
printf("%d",a);
return 0;

}

</syntaxhighlight>

Solution[edit]

<syntaxhighlight lang="c"> int *a; </syntaxhighlight> This declares a as an integer pointer, meaning a can point to a memory address which contains an int

<syntaxhighlight lang="c">

  • a = 5;

</syntaxhighlight> This makes the content of the address pointed to by a 5. But a is not pointing to any valid address (int *a assigns garbage value to a) and hence this assignment can cause segmentation fault



blog comments powered by Disqus