What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes?

#include<stdio.h>

int main()
{
    int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};
    printf("%u, %u\n", a+1, &a+1);
    return 0;
}

Since C is a machine dependent language sizeof(int) may return different values.

The output for the above program will be cd in Windows (Turbo C) and gh in Linux (GCC).

To understand it better, compile and execute the above program in Windows (with Turbo C compiler) and in Linux (GCC compiler).

Step 1: int x = 3; here variable x is an integer type and initialized to '3'.
Step 2: float y = 3.0; here variable y is an float type and initialized to '3.0'
Step 3: if(x == y) here we are comparing if(3 == 3.0) hence this condition is satisfied.
Hence it prints "x and y are equal".

Step 1: int i=1; The variable i is declared as an integer type and initialized to 1(one).

Step 2: if(!i) Here the !(NOT) operator reverts the i value 1 to 0. Hence the if(0) condition fails. So it goes to else part.

Step 3: else { i=0; In the else part variable i is assigned to value 0(zero).

Step 4: printf("C-Program"); It prints the "C-program".

Step 5: main(); Here we are calling the main() function.

After calling the function, the program repeats from step 1 to step 5 infinitely.

Hence it prints "C-Program" infinitely.

A label is used as the target of a goto statement, and that label must be within the same function as the goto statement.

Syntax: goto <identifier> ;
Control is unconditionally transferred to the location of a local label specified by <identifier>.
Example:

#include <stdio.h>
int main()
{
    int i=1;
    while(i>0)
    {
        printf("%d", i++);
        if(i==5)
          goto mylabel;
    }
    mylabel:
    return 0;
}
 

Output: 1,2,3,4

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.

Step 1:

printf("%d, %d, %d", sizeof(3.0f), sizeof('3'), sizeof(3.0));

The sizeof function returns the size of the given expression.

sizeof(3.0f) is a floating point constant. The size of float is 4 bytes

sizeof('3') It converts '3' in to ASCII value.. The size of int is 2 bytes

sizeof(3.0) is a double constant. The size of double is 8 bytes

Hence the output of the program is 4,2,8

Note: The above program may produce different output in other platform due to the platform dependency of C compiler.

In Turbo C, 4 2 8. But in GCC, the output will be 4 4 8.

Step 1: int k=1; The variable k is declared as an integer type and initialized to '1'.

Step 2: printf("%d == 1 is" "%s\n", k, k==1?"TRUE":"FALSE"); becomes

=> k==1?"TRUE":"FALSE"

=> 1==1?"TRUE":"FALSE"

=> "TRUE"

Therefore the output of the program is 1 == 1 is TRUE

Step 1: int fun(int); Here we declare the prototype of the function fun().

Step 2: int i = fun(10); The variable i is declared as an integer type and the result of the fun(10) will be stored in the variable i.

Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value return(i++). It returns 10. because i++ is the post-increement operator.

Step 4: Then the control back to the main function and the value 10 is assigned to variable i.

Step 5: printf("%d\n", --i); Here --i denoted pre-increement. Hence it prints the value 9.

Bit field type must be signed int or unsigned int.

The char type: char scheme:4; is also a valid statement.

When an automatic structure is partially initialized remaining elements are initialized to 0(zero).