Warning: implode(): Invalid arguments passed in /www/wwwroot/jobquiz.info/mdiscuss.php on line 336 What will be the output of the program? #include<stdio.h> int main() { const int x=5; const int ptrx; ptrx = &x; ptrx = 10; printf("%d\n", x); return 0; } ?->(Show Answer!)
1. What will be the output of the program? #include<stdio.h> int main() { const int x=5; const int ptrx; ptrx = &x; ptrx = 10; printf("%d\n", x); return 0; }
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By: guest on 01 Jun 2017 06.01 pm
Step 1: const int x=5; The constant variable x is declared as an integer data type and initialized with value '5'. Step 2: const int *ptrx; The constant variable ptrx is declared as an integer pointer. Step 3: ptrx = &x; The address of the constant variable x is assigned to integer pointer variable ptrx. Step 4: *ptrx = 10; Here we are indirectly trying to change the value of the constant vaiable x. This will result in an error. To change the value of const variable x we have to use *(int *)&x = 10;