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C Programming :: Functions

Home > C Programming > Functions > General Questions

1. What is the output of the following problem ?

#include
int main()
{
int j,ans;
j = 4;
ans = count(4);
printf("%d\n",ans);
return 0;
}
int count(int i)
{
if ( i < 0)
return(i);
else
return( count(i-2) + count(i-1));
}

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Answer: Option A

Explanation:

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2. What is the output of the following problem ?

#include
int main()
{
int j;
for(j=0;j<3;j++)
foo();
return 0;
}
foo() {
static int i = 10;
i+=10;
printf("%d",i);
}

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Answer: Option B

Explanation:

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3. What is the output of the following problem ?

#include

int main() {
int factorial(int n);
int i,ans;
ans = factorial(5);
printf("\nFactorial by recursion = %d\n", ans);
return 0;
}
int factorial(int n)
{
if (n <= 1)
return (1);
else
return ( n * factorial(n-1));
}

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Answer: Option A

Explanation:

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4. What is the output of the following 'C' program?

aaa() {
printf("hi");
}
bbb() {
printf("hello");
}
ccc() {
printf("bye");
}
void main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}

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Answer: Option B

Explanation:

ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

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5. What is the output of the following 'C' program?

void main(){
int i=3,val;
val=f(i)+ +f(i=1)+ +f(i-1);
printf("%d",val);
}
int f(int num){
return num*5;
}

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Answer: Option A

Explanation:

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Answer: Option C

Explanation:

Function names are just addresses (just like array names are addresses). main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers

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Capgemini 

7. int main()
{
show();
return 0;
}
void show()
{
printf("I'm the greatest");
}

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Answer: Option A

Explanation:

When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.

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Capgemini 

8. #include
int main()
{
int i=_l_abc(10);
printf("%d\n",--i);
return 0;
}
int _l_abc(int i)
{
return(i++);
}

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Answer: Option C

Explanation:

return(i++) it will first return i and then increments. i.e. 10 will be returned.

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9. #include
int main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}

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Answer: Option A

Explanation:

Labels have functions scope, in other words the scope of the labels is limited to functions. The label 'here' is available in function fun() Hence it is not visible in function main.

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10. #include
int AX = 0;
int main()
{
int i=0;
i = abc();
printf("%d",i);
return 0;
}
abc()
{
AX = 1000;
return AX;
}

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Answer: Option B

Explanation:

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