GATE CS 2015 Set 3 — Question 11
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Programming & Data Structures → C Programming → Arrays & Strings in C
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Question
Consider the following C program segment.
What will be printed by the program?
#include <stdio.h>
int main()
{
char s1[7] = "1234", *p;
p = s1 + 2;
*p = '0';
printf("%s", s1);
return 0;
}
What will be printed by the program?
Correct answer
(C) (C) 1204
Solution
Let's trace the execution of the C program:
1.
char s1[7] = "1234";- An array
s1of 7 characters is declared and initialized with the string "1234". - In memory,
s1will be{'1', '2', '3', '4', '\0', ?, ?}. The\0is the null terminator automatically added after "1234". The last two elements are uninitialized. - The indices are
s1[0] = '1',s1[1] = '2',s1[2] = '3',s1[3] = '4',s1[4] = '\0'.
char *p;- A character pointer
pis declared.
p = s1 + 2;- The pointer
pis made to point to the memory location ofs1[2]. So,pnow points to the character'3'.
*p = '0';- The character at the memory location pointed to by
pis changed to'0'. Sinceppoints tos1[2],s1[2]is updated from'3'to'0'. - The
s1array now looks like:{'1', '2', '0', '4', '\0', ?, ?}.
printf("%s", s1);- This statement prints the string starting from the address of
s1until a null terminator (\0) is encountered. - It will print
s1[0],s1[1],s1[2],s1[3], and then stop becauses1[4]is\0. - The characters printed will be
'1','2','0','4'.
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