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Do you have any idea what is the output of this program?

#include<stdio.h>
#include<stdlib.h>

int main()
{
int ptr;
ptr = 10;
ptr = 20;
printf("%dn",ptr);
return 0;
}
a) 10
b) 20
c) segmentation fault
d) none of the mentioned

Answer:

c) segmentation fault
Explanation:
The segmentation fault occurs because memory for the pointer has not been allocated in this program.
Output:
[root@localhost google]# gcc -o san san.c
[root@localhost google]# ./san
Segmentation fault (core dumped)
[root@localhost google]#

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What is the output of this program? #include<stdio.h> #include<stdlib.h> int main() { int ptr1, ptr2; ptr1 = malloc(4); ptr1 = 10; ptr2 = free(ptr1); printf("%dn",*ptr2); return 0; } a) 10 b) it will print the address stored in ptr1 c) it will print the address stored in ptr2 d) it will give an error	What is the output of this program? #include<stdio.h> int main() { int ptr; ptr = (int )calloc(1,sizeof(int)); ptr = 10; printf("%dn",ptr); return 0; } a) 0 b) -1 c) 10 d) none of the mentioned

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Do you have any idea what is the output of this program? #include<stdio.h> #include<stdlib.h> int main() { int *ptr; *ptr = 10; *ptr = 20; printf("%dn",*ptr); return 0; } a) 10 b) 20 c) segmentation fault d) none of the mentioned

Answer:

Do you have any idea what is the output of this program?

#include<stdio.h>
#include<stdlib.h>

int main()
{
int ptr;
ptr = 10;
ptr = 20;
printf("%dn",ptr);
return 0;
}
a) 10
b) 20
c) segmentation fault
d) none of the mentioned