What does the following program print ?

$void f(intp, intq){p=q;*p =2;}int i = 0, j = 1;int main( ){f(&i, &j);print f("%d%d \n", i, j);getchar( );return 0;}$

2, 2

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0, 2

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2, 1

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0, 1

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Solution

The correct option is B 0, 2
option (d)

$Step 1: initially 'p' has address of 'i' and 'q' has address of 'j' Step 2: p =q means Content of 'p' is updated with content of 'q' i.e. p has address of 'j' Step 3: *p=2; means content of 'j is updated with 2 Step 4: The output is 0, 2$

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What does the following program print ? void f(int*p, int*q){p=q;*p =2;}int i = 0, j = 1;int main( ){f(&i, &j);print f("%d%d \n", i, j);getchar( );return 0;}

The correct option is B 0, 2option (d) Step 1: initially 'p' has address of 'i' and 'q' has address of 'j'Step 2: p =q means Content of 'p' is updated with content of 'q' i.e. phas address of 'j'Step 3: *p=2; means content of 'j is updated with 2Step 4: The output is 0, 2

What does the following program print ?

$void f(intp, intq){p=q;*p =2;}int i = 0, j = 1;int main( ){f(&i, &j);print f("%d%d \n", i, j);getchar( );return 0;}$

The correct option is B 0, 2
option (d)

$Step 1: initially 'p' has address of 'i' and 'q' has address of 'j' Step 2: p =q means Content of 'p' is updated with content of 'q' i.e. p has address of 'j' Step 3: *p=2; means content of 'j is updated with 2 Step 4: The output is 0, 2$