What will be the output of the program (Turbo C in 16 bit platform DOS)?

#include<stdio.h>
#include<string.h>

int main()
{
    char *str1 = "India";
    char *str2 = "CURIOUSTAB";
    char *str3;
    str3 = strcat(str1, str2);
    printf("%s %s\n", str3, str1);
    return 0;
}

typedef long a;
extern int a c; while compiling this statement becomes extern int long c;. This will result in to "Declaration syntax error".

typedef long a;
extern a int c; while compiling this statement becomes extern long int c;. This will result in to "Too many types in declaration error".

typedef long a;
extern a c; while compiling this statement becomes extern long c;. This is a valid c declaration statement. It says variable c is long data type and defined in some other file or module.

So, Option C is the correct answer.

int length; denotes that variable length is int(integer) data type.

char int; here int is a keyword cannot be used a variable name.

int long; here long is a keyword cannot be used a variable name.

float double; here double is a keyword cannot be used a variable name.

So, the answer is int length;(Option A).

Because, comma operator used in the expression i (1, 2, 3, 4, 5). The comma operator has left-right associativity. The left operand is always evaluated first, and the result of evaluation is discarded before the right operand is evaluated. In this expression 5 is the right most operand, hence after evaluating expression (1, 2, 3, 4, 5) the result is 5, which on adding to i results into 7.

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i && ++j && ++k;
becomes m = -2 && 3 && 1;
becomes m = TRUE && TRUE; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j,k are increemented by '1'(one).

Hence the output is "-2, 3, 1, 1".

Step 1: int i; The variable i is declared as an global and integer type.

Step 2: int fun1(int); This prototype tells the compiler that the fun1() accepts the one integer parameter and returns the integer value.

Step 3: int fun2(int); This prototype tells the compiler that the fun2() accepts the one integer parameter and returns the integer value.

Step 4: extern int j; Inside the main function, the extern variable j is declared and defined in another source file.

Step 5: int i=3; The local variable i is defines as an integer type and initialized to 3.

Step 6: fun1(i); The fun1(i) increements the given value of variable i prints it. Here fun1(i) becomes fun1(3) hence it prints '4' then the control is given back to the main function.

Step 7: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.

Step 8: fun2(i); The fun2(i) increements the given value of variable i prints it. Here fun2(i) becomes fun2(3) hence it prints '4' then the control is given back to the main function.

Step 9: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.

Hence the output is "4 3 4 3".

If you compile and execute this program in windows platform with Turbo C, it will give "lice ice ce e".

It may give different output in other platforms (depends upon compiler and machine). The online C compiler given in this site will give the Option C as output (it runs on Linux platform).

The macro PRINT(int) print("%d,", int); prints the given variable value in an integer format.

Step 1: int x=2, y=3, z=4; The variable x, y, z are declared as an integer type and initialized to 2, 3, 4 respectively.

Step 2: PRINT(x); becomes printf("int=%d,",x). Hence it prints 'int=2'.

Step 3: PRINT(y); becomes printf("int=%d,",y). Hence it prints 'int=3'.

Step 4: PRINT(z); becomes printf("int=%d,",z). Hence it prints 'int=4'.

Hence the output of the program is int=2, int=3, int=4.

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.