Study the code:
void show()
main()
{
show();
}
void show (char *s)
{
printf("%sn",s);
}
What will happen if it is compiled & run on an ANSI C
Compiler?
A)It will compile & nothing will be printed when it is
executed
B)it will compile but not link
C)the compiler will generate an error
D)the compiler will generate a warning
- 4 Answers
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Answers were Sorted based on User's Feedback
Answer / manjeet
The compiler will generate an error because the is no ";"
semicolon after the prototype declaration of show function
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / manjeet
Then answer wll be (B) will compile but not link
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / subbu[iit kgp]
if it is assumed that there is a semicolon after the
function prototype declaration of show(), then answer is A.
| Is This Answer Correct ? | 1 Yes | 2 No |
Answer / vignesh1988i
according to my compailer it will have an error "function
called incorrectly"
| Is This Answer Correct ? | 0 Yes | 1 No |
what wud be the output? main() { char *str[]={ "MANISH" "KUMAR" "CHOUDHARY" }; printf("\nstring1=%s",str[0]); printf("\nstring2=%s",str[1]); printf("\nstring3=%s",str[2]); a)string1=Manish string2=Kumar string3=Choudhary b)string1=Manish string2=Manish string3=Manish c)string1=Manish Kumar Choudhary string2=(null) string3=(null) d)Compiler error
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