What will be the output of the program?

#include<stdio.h>

int main()
{
    int i;
    char c;
    for(i=1; i<=5; i++)
    {
        scanf("%c", &c); /* given input is 'a' */
        printf("%c", c);
        ungetc(c, stdin);
    }
    return 0;
}

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.

In the program, printf() returns the number of charecters printed on the console i = printf("How r u\n"); This line prints "How r u" with a new line character and returns the length of string printed then assign it to variable i.
So i = 8 (length of '\n' is 1).

i = printf("%d\n", i); In the previous step the value of i is 8. So it prints "8" with a new line character and returns the length of string printed then assign it to variable i. So i = 2 (length of '\n' is 1).

printf("%d\n", i); In the previous step the value of i is 2. So it prints "2".

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

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'.

scanf() returns the number of variables to which you are provding the input. i = scanf("%d %d", &i, &i); Here Scanf() returns 2. So i = 2.

printf("%d\n", i); Here it prints 2.

Function atoi() converts the string to integer.
Function atof() converts the string to float.

result1 = result1+atoi(i);
Here result1 = 10 + atoi(55.555);
result1 = 10 + 55;
result1 = 65;

result2 = result2+atof(i);
Here result2 = 11.111 + atof(55.555);
result2 = 11.111 + 55.555000;
result2 = 66.666000;
So the output is "65, 66.666000" .

Step 1: x=y=z=1; here the variables x ,y, z are initialized to value '1'.

Step 2: z = ++x || ++y && ++z; becomes z = ( (++x) || (++y && ++z) ). Here ++x becomes 2. So there is no need to check the other side because ||(Logical OR) condition is satisfied.(z = (2 || ++y && ++z)). There is no need to process ++y && ++z. Hence it returns '1'. So the value of variable z is '1'

Step 3: printf("x=%d, y=%d, z=%d\n", x, y, z); It prints "x=2, y=1, z=1". here x is increemented in previous step. y and z are not increemented.

printf() returns the number of charecters printed on the console.

Step 1: if(ch = printf("")) here printf() does not print anything, so it returns '0'(zero).
Step 2: if(ch = 0) here variable ch has the value '0'(zero).
Step 3: if(0) Hence the if condition is not satisfied. So it prints the else statements.
Hence the output is "It doesn't matters".

Note: Compiler shows a warning "possibly incorrect assinment".

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)

Here unsigned int size is 2 bytes. It varies from 0,1,2,3, ... to 65535.

Step 1:unsigned int i = 65536; here variable i becomes '0'(zero). because unsigned int varies from 0 to 65535.

Step 2: while(i != 0) this statement becomes while(0 != 0). Hence the while(FALSE) condition is not satisfied. So, the inside the statements of while loop will not get executed.

Hence there is no output.

Note: Don't forget that the size of int should be 2 bytes. If you run the above program in GCC it may run infinite loop, because in Linux platform the size of the integer is 4 bytes.