Determine Output:
void main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s", string);
}
void main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s", string);
}A. will print Hello World
B. Compiler Error
C. Can't Say
D. None of These
Answer: Option B
Solution(By Examveda Team)
Type mismatch in redeclaration of function display.
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.
This Question Belongs to C Program >> C Miscellaneous
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Comments ( 5 )
Determine output:
void main()
{
int const *p=5;
printf("%d", ++(*p));
}
A. 6
B. 5
C. Garbage Value
D. Compiler Error
A. mmm nnn aaa
B. mmmm nnnn aaaa
C. Compiler Error
D. None of These
A. I hate Examveda
B. I love Examveda
C. Error
D. None of These
Determine Output:
void main()
{
static int var = 5;
printf("%d ", var--);
if(var)
main();
}
A. 5 5 5 5 5
B. 5 4 3 2 1
C. Infinite Loop
D. None of These
it prints Hello World
But it prints Hello world.
conflict definitely occurs with the function display. so the output is not hello world or what so ever
output is Hello world
no output is : - Hello World