c_programming

Question

What will be the output of the program?

#include<stdio.h>
int main()
{
    const c = -11;
    const int d = 34;
    printf("%d, %d\n", c, d);
    return 0;
}

    A.    Error    
    B.    -11, 34
    C.    11, 34
    D.    None of these

is it const c = -11; correct definition?

Answer 1

http://stackoverflow.com/questions/11064292/declaring-variables-without-any-data-type-in-c

A variable declared without an explicit type name is assumed to be of type int.

It's a valid C program and the answer is B.

Comments

@Arjun Sir verify this.

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A variable declared without an explicit type name is assumed to be of type int.

This was quoted from the official language specification. 

http://ideone.com/6e8t2k

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is it always be "int" ?

like if instead of  printf("%d, %d\n", c, d); if we hve

 printf("%c, %d\n", c, d); then i think it will give error becoz c=-11

if it is c=11 then it print ASCII value of 11

correct?

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-11 will be converted to signed char. So it will become 245 and the character corresponding to ASCII value 245 will be printed. http://www.asciitable.com/

http://stackoverflow.com/questions/13161199/ascii-table-negative-value

Answer 2

Output would be -11,34 because in c if we write const c..this means ci by default int...

Answer 3

const c = -11; is not a correct definition in standard C; you must specify a type explicitly.

Answer 4

B is correct