Simple C Language placement Question and answer

96) void main()

{

static int i=5;

if(--i)

{

main();

printf("%d ",i);

}

}

Answer:-

 0 0 0 0

Explanation:

The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

2)

void main()

{

int k=ret(sizeof(float));

printf("\n here value is %d",++k);

}

int ret(int ret)

{

ret += 2.5;

return(ret);

}

Answer:-

 Here value is 7

Explanation:

The int ret(int ret), ie., the function name and the argument name can be the same.

Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed,  after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.

 

3)

void main()

{

char a[]="12345\0";

int i=strlen(a);

printf("here in 3 %d\n",++i);

}

Answer: 

here in 3 6

Explanation:

The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.

4)

void main()

{

unsigned giveit=-1;

int gotit;

printf("%u ",++giveit);

printf("%u \n",gotit=--giveit);

}

Answer:

 0 65535

5)

void main()

{

int i;

char a[ ]="\0";

if(printf("%s\n",a))

printf("Ok here \n");

else

printf("Forget it\n");

}

Answer:-

 Ok here 

Explanation:

Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.

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