Study the code: void show() main() { show(); } void show (char *s) { pri

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

Question Posted / guest

4 Answers
6087 Views
Accenture, I also Faced
E-Mail Answers

Answers were Sorted based on User's Feedback

Study the code: void show() main() { show(); } void show (char *s) { printf("%sn",..

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

Study the code: void show() main() { show(); } void show (char *s) { printf("%sn",..

Answer / manjeet

Then answer wll be (B) will compile but not link

Is This Answer Correct ?

1 Yes

1 No

Study the code: void show() main() { show(); } void show (char *s) { printf("%sn",..

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

Study the code: void show() main() { show(); } void show (char *s) { printf("%sn",..

Answer / vignesh1988i

according to my compailer it will have an error "function
called incorrectly"

Is This Answer Correct ?

0 Yes

1 No

Post New Answer