#include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
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Answer / surenda pal singh chouhan
Compiler Error: Constant expression required in function
main.
Explanation:
The case statement can have only constant expressions (this
implies that we cannot use variable names directly so an
error).
Note:
Enumerated types can be used in case statements.
| Is This Answer Correct ? | 10 Yes | 2 No |
Answer / subha raman
yeah..itz mainly syntax error..
there shud not be any declaration of variables..in
case "j"..it must be "case 2"only..
| Is This Answer Correct ? | 6 Yes | 1 No |
Answer / pravin
if we use the single qutoes' 'at 1 and j the rest of
program is right because we already decleared the value of i
and i=1 .
so output should be "GOOD" only. (without" ")
| Is This Answer Correct ? | 3 Yes | 1 No |
Answer / moolshankershukla
#include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
given above program is wrong only we can one changes and
will be run .
correct program is:
#include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case 2: printf("BAD");
break;
}
}
| Is This Answer Correct ? | 2 Yes | 0 No |
Compiler Error: Constant expression required in function
main
instead of using j u we can use case 'j' .
this is correct answer
| Is This Answer Correct ? | 5 Yes | 5 No |
stripos — Find position of first occurrence of a case- insensitive string int stripos ( char* haystack, char* needle, int offset ) Returns the numeric position of the first occurrence of needle in the haystack string. Note that the needle may be a string of one or more characters. If needle is not found, stripos() will return -1. The function should not make use of any C library function calls.
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C 
