How many bytes of memory will the following code reserve?

#include<stdio.h>
#include<stdlib.h>

int main()
{
    int *p;
    p = (int *)malloc(256 * 256);
    if(p == NULL)
        printf("Allocation failed");
    return 0;
}

Both ceil() and floor() return the integer found as a double.

floor(2.5) returns the largest integral value(round down) that is not greater than 2.5. So output is 2.000000.

ceil(2.5) returns 3, while converting the double to int it returns '0'.
So, the output is '2.000000, 0'.

The gcvt() function converts a floating-point number to a string. It converts given value to a null-terminated string.

#include <stdlib.h>
#include <stdio.h>

int main(void)
{
	char str[25];
	double num;
	int sig = 5; /* significant digits */

	/* a regular number */
	num = 9.876;
	gcvt(num, sig, str);
	printf("string = %s\n", str);

	/* a negative number */
	num = -123.4567;
	gcvt(num, sig, str);
	printf("string = %s\n", str);

	/* scientific notation */
	num = 0.678e5;
	gcvt(num, sig, str);
	printf("string = %s\n", str);

	return(0);
}

Output:
string = 9.876
string = -123.46
string = 67800

The ungetc() function pushes the character c back onto the named input stream, which must be open for reading.

This character will be returned on the next call to getc or fread for that stream.

One character can be pushed back in all situations.

A second call to ungetc without a call to getc will force the previous character to be forgotten.

Step 1: int fun(int); This is prototype of function fun(). It tells the compiler that the function fun() accept one integer parameter and returns an integer value.

Step 2: int i=3; The variable i is declared as an integer type and initialized to value 3.

Step 3: fun(i=fun(fun(i)));. The function fun(i) increements the value of i by 1(one) and return it.

Lets go step by step,

=> fun(i) becomes fun(3) is called and it returns 4.

=> i = fun(fun(i)) becomes i = fun(4) is called and it returns 5 and stored in variable i.(i=5)

=> fun(i=fun(fun(i))); becomes fun(5); is called and it return 6 and nowhere the return value is stored.

Step 4: printf("%d\n", i); It prints the value of variable i.(5)

Hence the output is '5'.

for(i<=5 && i>=-1; ++i; i>0) so expression i<=5 && i>=-1 initializes for loop. expression ++i is the loop condition. expression i>0 is the increment expression.

In for( i <= 5 && i >= -1; ++i; i>0) expression i<=5 && i>=-1 evaluates to one.

Loop condition always get evaluated to true. Also at this point it increases i by one.

An increment_expression i>0 has no effect on value of i.so for loop get executed till the limit of integer (ie. 65535)

/* Example for ceil() and floor() functions: */

#include<stdio.h>
#include<math.h>

int main()
{
    printf("\n Result : %f" , ceil(1.44) );
    printf("\n Result : %f" , ceil(1.66) );
 
    printf("\n Result : %f" , floor(1.44) );    
    printf("\n Result : %f" , floor(1.66) );

    return 0;
}
// Output:
// Result : 2.000000
// Result : 2.000000
// Result : 1.000000
// Result : 1.000000

The keyword continue cannot be used in switch case. It must be used in for or while or do while loop. If there is any looping statement in switch case then we can use continue.