Discussion
What will be the output of the following C code?
#include <stdio.h>
int main()
{
char *str = "hello, world";
char str1[15] = "hello wo 9";
strcpy(str, str1);
printf("%s", str1);
}
Answer: B
Correct answer: Segmentation Fault
str points to a string literal, which almost every compiler places in read-only memory. strcpy then tries to write over it, and the program is killed by the memory protection. Declaring char str[20] = "hello, world"; would give a writable copy.
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