CODEWITHSUNDEEP
c
Chris
Chris

Reputation: 672

'Char' is promoted to 'int' when passed through in C

I have a question. I have line in my program:

sprintf(buff,"Nieznany znak: %d",(char)va_arg(x, const char)); break;

Why after compile I have an error:

error.c: In function 'error':
error.c:30:46: warning: 'char' is promoted to 'int' when passed through '...' [e
nabled by default]
  case 0xfe: sprintf(buff,"Nieznany znak: %d",(char)va_arg(x, const char)); brea
k;
                                              ^
error.c:30:46: note: (so you should pass 'int' not 'char' to 'va_arg')
error.c:30:46: note: if this code is reached, the program will abort

In my opinion everything is okay, why I can't do this like that?

Upvotes: 8

Views: 7177

Answers (2)

epic
epic

Reputation: 1443

As noted in this link

"This noncompliant code example attempts to read a variadic argument of type unsigned char with va_arg(). However, when a value of type unsigned char is passed to a variadic function, the value undergoes default argument promotions, resulting in a value of type int being passed."

Solution:

Access variadic function with type int and cast it back to unsigned char (or char)

unsigned char c = (unsigned char) va_arg(ap, int);

Upvotes: 3

The Paramagnetic Croissant
The Paramagnetic Croissant

Reputation: 9343

The compiler just told you why:

'char' is promoted to 'int' when passed through '...'

According to the C language specification, usual arithmetic conversions are applied to the unnamed arguments of variadic functions. Hence, any integer type shorter than int (e. g. bool, char and short) are implicitly converted int; float is promoted to double, etc.

Do what the compiler asks you for (use va_arg() with int as its second parameter), else your code will invoke undefined behavior.

Upvotes: 18

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