C Programming - Variable Number of Arguments - Discussion
Discussion Forum : Variable Number of Arguments - Point Out Errors (Q.No. 2)
2.
Point out the error if any in the following program (Turbo C).
#include<stdio.h>
#include<stdarg.h>
void display(int num, ...);
int main()
{
display(4, 'A', 'a', 'b', 'c');
return 0;
}
void display(int num, ...)
{
char c; int j;
va_list ptr;
va_start(ptr, num);
for(j=1; j<=num; j++)
{
c = va_arg(ptr, char);
printf("%c", c);
}
}
Discussion:
15 comments Page 2 of 2.
Arunmane said:
10 years ago
I think, it is undefined behaviour. Because in the va_arg will return character means, char can not be copied into char variable.
Example: Char c = 'A'; // this is correct way.
But in the va_arg will return char value not its ASCII, so it will be like this.
Char c = A // This is wrong. So it is undefined behaviour.
Example: Char c = 'A'; // this is correct way.
But in the va_arg will return char value not its ASCII, so it will be like this.
Char c = A // This is wrong. So it is undefined behaviour.
Robert said:
9 years ago
Guys, watch out for some warning that will give runtime error with GCC compiler:
\Untitled1.c|13|warning: 'char' is promoted to 'int' when passed through '...'|
\\Untitled1.c|13|note: (so you should pass 'int' not 'char' to 'va_arg')|
\Untitled1.c|13|note: if this code is reached, the program will abort|
Normally the remedy for this should be to pass instead of : c = va_arg(ptr, char)
c = va_arg(ptr, int);
\Untitled1.c|13|warning: 'char' is promoted to 'int' when passed through '...'|
\\Untitled1.c|13|note: (so you should pass 'int' not 'char' to 'va_arg')|
\Untitled1.c|13|note: if this code is reached, the program will abort|
Normally the remedy for this should be to pass instead of : c = va_arg(ptr, char)
c = va_arg(ptr, int);
Pooja said:
8 years ago
What is va_start?
Sowjanya said:
8 years ago
Please give clear explanation about this question.
V.Nandhini said:
6 years ago
I am not understanding it. Please, anyone, help me to get it.
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