CODEWITHSUNDEEP
c++char
oldMCdonald
oldMCdonald

Reputation: 55

what is the value of a multi-character character constant variable

I know multi-character character constant stated as int. and I know that the value of it is compiler dependent. but my question is when I store a multi-character character constant in a char variable it will behave in a different way.

#include <iostream>

int main() {
    std::cout << 'asb';
    return 0;
}

output: 6386530

#include <iostream>

int main() {
    char a = 'asb';
    std::cout << a;
    return 0;   
}

output: b

Upvotes: 3

Views: 352

Answers (3)

user30701
user30701

Reputation: 308

Case 1 : You are getting 'a'*256²+'s'*256+'b' = 6386530 because 'a' = 97, 's' = 115, 'b' = 98

cf. Ascii table

'asb' is interpreted as an integer. 

typeid('asb').name()[0] == 'i' && sizeof('asd') == 4;

An integer is 32 bits, and you can store 'asb' (24bits) in an integer. That's why std::cout interprets it as an integer and display 6386530

Note that also:

  typeid('xxxxabcd').name()[0] == 'i' && sizeof('xxxxabcd') == 4;

but 'xxxxabcd' is represented by 64-bits, so 32-bits are lost. 

std::cout << 'xxxxabcd';
std::cout << 'abcd';

would print the same thing.

Case 2 : 'asb' is interpreted as an integer and you cast it into a char (8-bits). As @BenjaminJones pointed out, only the last 8-bits (98=='b') are saved. And std::cout interprets it as a char so it displays 'b'.

Anyway, both case provokes compilation warning such as :

warning: multi-character character constant [-Wmultichar] 
warning: multi-character character constant [-Wmultichar] In function 'int main()'
warning: overflow in implicit constant conversion [-Woverflow]

I guess the behavior depends on the compiler.

Upvotes: 3

Ben Jones
Ben Jones

Reputation: 629

In your second example, the literal is an int that gets converted to a char. Therefore, there are a few implementation specific issues here:

  • Interpretation of a multi-character literal
  • Whether char is signed or unsigned
  • Conversion from an int to an signed char (if char is signed)

It appears that what is happening is that 'asb' is getting interpreted as 6386530, and then truncated according to the rules for conversion from int to an unsigned char. In other words, 6386530 % 256 == 97 == 'b'.

Upvotes: 1

StoryTeller - Unslander Monica
StoryTeller - Unslander Monica

Reputation: 170203

You don't store the value of the multi-char constant as is. You convert the value into another value that fits in the range of a char. Since you are also printing the value via entirely different overloads of operator<< (the one for char instead of the one for int), it stands to reason the output would be different on account of that too.

Upvotes: 1

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