Scanning the Array of pointers issue

Question

#include<stdio.h>
#define SIZE 3
int main() 
{
char *arr[5];
for(i = 0; i < 5; i++) {
arr[i] = (char *)malloc(SIZE * sizeof(char));
scanf("%s",arr[i]);
}
for(i = 0; i < 5; i++) {
printf("%s",arr[i]);
}
return 0;
}

In the above code i am scanning more than 3(SIZE) characters but it is working fine.what is the issue?

Undefined behavior. It includes seemingly sensible behavior. — R Sahu, Feb 20 '15 at 05:02
When you call `malloc(3)`, it is possible that you get the equivalent of `malloc(16)` because the standard library implementation doesn't want to divide memory into such small chunks. But another implementation might give you just `3` bytes. In the first case, you will get away with reading upto 15 characters while in the second case, you won't. — R Sahu, Feb 20 '15 at 05:07

score 0 · Answer 1 · edited May 23 '17 at 11:43

0

It is undefined behavior. In linux, doesn't strictly follows the array out of bounds. If there is a free memory it will allow you to access that memory. Refer this link.

In windows, it will strictly follow the array out of bounds. In linux, we have make sure that we are not accessing the out of memory.

edited May 23 '17 at 11:43

Community

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answered Feb 20 '15 at 05:06

Karthikeyan.R.S

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score 0 · Answer 2 · answered Feb 20 '15 at 05:09

Yes it works but it is not safe. arr[i] is just an address in RAM when you do scanf("%s",arr[i]); The characters read will be stored sequentially into the adress starting from pointed out by arr[i].

Lets say you had a memory of 100 chars when you did malloc you may get addresses 10, 20, 30, 40 and 50. Even when you read 5 characters into each of these starting addresses it works. But in case if malloac returns address pointers to 10, 13, 16, 19 and 22. When you do scanf("%s",arr[0]); with 5 chars say 'abcde' the string will be stored in locations from 10 to 15 (one for \0). When you do scanf("%s",arr[1]); with input 'fghij' it will start writing from location 13. Now if you do printf on arr[0] you will get abcfghij.

This behavior depends on the compiler.

Scanning the Array of pointers issue

2 Answers2