Segmentation Fault after 5th scanf

Question

This code gives a Segmentation Fault at the exact 5th line of input. If n<5 or in the debugger everything is fine.

scanf("%d %d",&n,&x);
m=(int**)malloc(n*sizeof(int));
for(i=0;i<n;i++)
{
     m[i]=(int*)malloc(2*sizeof(int));
 }

 for(i=0;i<n;i++)
{
  scanf("%d %d",&m[i][0],&m[i][1]); 

 }`

Do not cast the result of `malloc` in C! – Spikatrix Feb 07 '15 at 16:43 — Spikatrix, Feb 07 '15 at 16:43

pmg · Accepted Answer · 2015-02-08T08:13:55.237

4

assumes sizeof (int) and sizeof (int*) are the same

m=(int**)malloc(n*sizeof(int));

Try this

m = malloc(n * sizeof *m);

edited Feb 08 '15 at 08:13

answered Feb 07 '15 at 16:38

pmg

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score 0 · Answer 2 · answered Feb 07 '15 at 16:42

0

Your mistake: You allocate memory for ints while you need int pointers

(As your m array is going to hold pointers for other int arrays)

mistake:

m=(int**)malloc(n*sizeof(int));

correct:

m=(int**)malloc(n*sizeof(int*));

answered Feb 07 '15 at 16:42

SomethingSomething

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Do not cast the result of `malloc` in C! – Spikatrix Feb 07 '15 at 16:43
Why? And if it was C++? What's the difference? – SomethingSomething Feb 07 '15 at 23:46
Thank you for that important comment! I've just seen this post: http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc – SomethingSomething Feb 07 '15 at 23:50

Segmentation Fault after 5th scanf

2 Answers2