String literals for array

Question

#include <stdio.h>

int main(void) {
    char a[] = "125"; // (int)1, (int)2, (int)5. But array 'a' has a type char. So int is in char. ???
    printf("%s", a);
}

In that code, each element of string literal has type int. But the array a has type char.

In C99 6.4.5 $2 fragment
The same considerations apply to each element of the sequence in a character string literal or a wide string literal as if it were in an integer character constant or a wide character constant

In C99 6.4.5 $5 fragment
For character string literals, the array elements have type char, and are initialized with the individual bytes of the multibyte character sequence

I think they are not compatible, it's a contradiction to me. What's wrong about my thought?

"Each element of string literal has type int." -- That is the part where you are wrong. A character literal has type int. A string literal has type char []. — DevSolar
– DevSolar, Commented Sep 28, 2020 at 7:41

Sourav Ghosh · Accepted Answer · 2020-09-28 07:53:51Z

2

No, this is a string literal.

Quoting C11, chapter 6.4.5, String Literals:

A character string literal is a sequence of zero or more multibyte characters enclosed in double-quotes, as in "xyz".[...]

To elaborate, the acceptable syntax for a string liteal is:

string-literal:
                 encoding-prefixopt " s-char-sequenceopt "
          encoding-prefix:
                 u8
                 u
                 U
                 L
          s-char-sequence:
                 s-char
                 s-char-sequence s-char
          s-char:
                 any member of the source character set except
                              the double-quote ", backslash \, or new-line character
                 escape-sequence

and then, the "source character set", referring (Chapter 5.2.1/P3)

Both the basic source and basic execution character sets shall have the following members: the 26 uppercase letters of the Latin alphabet

         A   B   C   D   E   F   G   H   I   J   K   L   M
         N   O   P   Q   R   S   T   U   V   W   X   Y   Z
the 26 lowercase letters of the Latin alphabet
         a   b   c   d   e   f   g   h   i   j   k   l   m
         n   o   p   q   r   s   t   u   v   w   x   y   z
the 10 decimal digits
         0   1   2   3   4   5   6   7   8   9
the following 29 graphic characters
         !   "   #   %   &   '   (   )   *   +   ,   -   .   /   :
         ;   <   =   >   ?   [   \   ]   ^   _   {   |   }   ~

So, a construct like "123" is a string literal, not individual integers held by/in char.

edited Sep 28, 2020 at 7:53

answered Sep 28, 2020 at 7:40

Sourav Ghosh

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2 Comments

op ol Over a year ago

Thanks, I misused the term character string literal. Then, in char a[] = "xyz", does each of three elements('x', 'y', 'z') have type int according to the standard? Because "xyz" is a character string literal and integer character constant has type int. Standard says that we regard each element of sequence in a character string literal as if it were in an integer character constant.

Ian Abbott Over a year ago

@opol char a[] = "xyz"; is equivalent to char a[] = { 'x', 'y', 'z', '\0' }; as long as x, y, and z are single-byte characters, and 'x', 'y','z' and '\0' do have type int. It is OK to initialize a char with an int value as long as the value can be represented by the type. That's good because C has no constants of type char, signed char, unsigned char, short or unsigned short.

tdao · Accepted Answer · 2020-09-28 07:42:02Z

1

char a[] = "125";

In that code, each element of string literal has type int. But the array a has type char.

No, the fact that it's a 5 does not mean it has to be an int. The type of which has to be determined by the context of where/how it is declared.

In your case that 5 is of type char because it is part of the string literal.

Also note that 5 can be one of any other types such as unsigned int, unsigned short, double, etc. So again you must look at how it's declared in the first place.

answered Sep 28, 2020 at 7:42

tdao

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