C | String | Question 1
Consider the following code. The function myStrcat concatenates two strings. It appends all characters of b to end of a. So the expected output is “Geeks Quiz”. The program compiles fine but produces segmentation fault when run.
#include <stdio.h> void myStrcat(char *a, char *b){ int m = strlen(a); int n = strlen(b); int i; for (i = 0; i <= n; i++) a[m+i] = b[i];} int main(){ char *str1 = "Geeks "; char *str2 = "Quiz"; myStrcat(str1, str2); printf("%s ", str1); return 0;} |
Which of the following changes can correct the program so that it prints “Geeks Quiz”?
(A) char *str1 = “Geeks “; can be changed to char str1[100] = “Geeks “;
(B) char *str1 = “Geeks “; can be changed to char str1[100] = “Geeks “; and a line a[m+n-1] = ‘\0’ is added at the end of myStrcat
(C) A line a[m+n-1] = ‘\0’ is added at the end of myStrcat
(D) A line ‘a = (char *)malloc(sizeof(char)*(strlen(a) + strlen(b) + 1)) is added at the beginning of myStrcat()
Answer: (A)
Explanation: In the above program “Geeks ” and “Quiz” stored in read-only location but the line a[m+i]=b[i] tries to write a read only memory. The problem can be solved by storing str in writable stack segment. For more details see following. https://www.geeksforgeeks.org/storage-for-strings-in-c/
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