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What will be the output of the program ? #include<stdio.h> void fun(int p); int main() { int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; int ptr; ptr = &a[0][0]; fun(&ptr); return 0; } void fun(int p) { printf("%d\n", p); } ?->(Show Answer!)
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1. What will be the output of the program ? #include<stdio.h> void fun(int p); int main() { int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; int ptr; ptr = &a[0][0]; fun(&ptr); return 0; } void fun(int p) { printf("%d\n", p); }





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  • By: guest on 01 Jun 2017 06.00 pm
    Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; The variable a is declared as an multidimensional integer array with size of 3 rows 4 columns. Step 2: int *ptr; The *ptr is a integer pointer variable. Step 3: ptr = &a[0][0]; Here we are assigning the base address of the array a to the pointer variable *ptr. Step 4: fun(&ptr); Now, the &ptr contains the base address of array a. Step 4: Inside the function fun(&ptr); The printf("%d\n", **p); prints the value '1'. because the *p contains the base address or the first element memory address of the array a (ie. a[0]) **p contains the value of *p memory location (ie. a[0]=1). Hence the output of the program is '1'
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