How exactly is this function an example of a char to int conversion?

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The book The C Programming Language by Kernighan and Ritchie, second edition states on page 43 in the chapter about Type Conversions:

Another example of char to int conversion is the function lower, which maps a single character to lower case for the ASCII character set. If the character is not an upper case letter, lower returns returns it unchanged.

/* lower: convert c to lower case; ASCII only */ int lower(int c) {     if (c >= 'A' && c <= 'Z')         return c + 'a' - 'A';     else         return c; } 

It isn't mentioned explicitly in the text so I'd like to make sure I understand it correctly: The conversion happens when you call the lower function with a variable of type char, doesn't it? Especially, the expression

c >= 'A' 

has nothing to do with a conversion from int to char since a character constant like 'A' is handled as an int internally from the start, isn't it? Edit: Or is this different (e.g. a character constant being treated as a char) for ANSI C, which the book covers?

 


K&R C is old. Really old. Many particulars of K&R C are no longer true in up-to-date standard C.

In stadard, up-to-date C11, there is no conversion to/from char in the function you posted:

/* lower: convert c to lower case; ASCII only */ int lower(int c) {     if (c >= 'A' && c <= 'Z')         return c + 'a' - 'A';     else         return c; } 

The function accepts int arguments as int c, and per 6.4.4.4 Character constants of the C standard, character literals are of type int.

Thus the entire lower function, as posted, under C11 deals entirely with int values.

The conversion, if any, is may be done when the function is called:

char upperA = 'A`;  // this will implicitly promote the upperA char // value to an int value char lowerA = lower( upperA ); 

Note that this is one of the differences between C and C++. In C++, character literals are of type char, not int.

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