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In which of the following case(s) is it possible to obtain different results for call-by-reference and call-by-name parameter passing?

  1. Passing an expression as a parameter
  2. Passing an array as a parameter
  3. Passing a pointer as a parameter
  4. Passing as array element as a parameter
in Compiler Design by (371 points) | 1.5k views
+2
0

ans will be D

1 Answer

+14 votes
Best answer

Answer A, D.

A is correct as call-by-name works like a macro and substitution happens only during use time. For example if we pass $2+3$ to the below function

int foo(int x)
{
    return x * x;
}

we get $2+3*2+3$ which will be $11$ due to the higher precedence for $*.$ But, call by reference will return $5*5 = 25.$ (For call by reference, when an expression is passed, a temporary variable is created and passed to the function) 

D is also correct: Passing an array element as a parameter

See the below example:

void m(int x,int y){
    for(int k = 0;k < 10;k++){
        y = 0; x++;
    }
}

int main(){
    int j; int A[10];
    j = 0;
    m(j,A[j]);
    return 0;
}


For the above example if we use 'Call by name' its initialize all the array elements with $0.$ But if we apply ' Call by Reference ' it will only initialize $A[0]$ with $0.$

by Active (2.4k points)
selected by
0

How to pass 2+3 as actual parameter? Pls explain using calling statement.

@mystylecse

+1

Interestingly the pass by value seems to be giving output as 25 and not 10.

Check ideone link.

#include <stdio.h>

int foo(int x)
{
    return x * x;
}

int main(void) {
	// your code goes here
	int a = foo(2+3);
	printf("a is %d\n",a);
	return 0;
}

 

+3
It's 25  because C does not support call by name.
+1

@Ekta07_GATE

You are right!

//Simulating pass by name in C using macros
#include <stdio.h> 
#define MULTIPLY(a, b) a*b 
int main() 
{ 
    // The macro is expanded as 2 + 3 * 3 + 5, not as 5*8 
    printf("%d", MULTIPLY(2+3, 3+5)); 
    return 0; 
}

Ideone link.

Source: geeksforgeeks example 4

 

0

@Arjun sir

i am getting a doubt with option b ie. Passing an array as parameter. Consider code below:

Let integer pointer takes 4 bytes and integer also takes 4 bytes.

int fun(int a[])

{

int n= sizeof(a)/sizeof(a[0]);

printf("%d",n);

}

int main()

{

int arr[10];

fun(arr);

return 0;

}

Call by name: 10

Call by reference: 1

Sir please clarify.

0
Please anyone clarify my doubt
Answer:

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