Programming Output

Question

Determine the output-

#include <stdio.h>
int main(void) 
{
	char *p="gateoverflow";
	*(p+5)='z';
	printf("%s",p);
	return 0;
}

Comments

gateozerflow

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Runtime Error

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i think 

gateozerflow

@GateSet, why Runtime error? any reason if you got.

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# include <stdio.h>
int main()
{
char s1[7] = "1234", *p;
p = s1 + 2;
*p = '0';
printf("%s", s1);  }

this is same as above right??

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@bad_engineer You are right. I modified your code to match mine. And it is compiling and giving output. I don't know why mine is giving runtime error.

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Segmentation fault

Answer 1

The declaration :

char *p  =  "gateoverflow" ;

so this suggests that "gateoverflow" is a string constant..And for string constant :

Base address can be changed but content cannot be changed.. 

And reverse happens for array declaration like :

char p[15] = "Gateoverflow" .Here base address cannot be changed but content can be changed..

Hence the above code results to runtime error..