char ch = 'a';
Here ch is a character variable, so it's size is one byte. 'a' is a character constant,so it's ASCII value will be stored which is 2 byte.But how could it possible to store a 2 byte value in an 1 byte variable ?
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How is the ASCII value of 'a' two bytes long? ASCII is a 7-bit character encoding standard. – dandan78 Oct 14 '10 at 13:24
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2Do not understand. ASCII is a set of 128 characters, so 7 bits should be sufficient to represent all ASCII chars. – Benoit Oct 14 '10 at 13:24
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But the size of a character constant is 2 byte or 4 byte ...so how could it possible to store it in 1 byte variable ? – Parikshita Oct 14 '10 at 13:27
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1Maybe Joel will be able to help you: http://www.joelonsoftware.com/articles/Unicode.html – ruslik Oct 14 '10 at 13:31
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1I'm wondering where you got that misinformation from? `sizeof(char)` is defined to be 1. – AndiDog Oct 14 '10 at 13:32
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-1 for bogus assumption with no explanation of where you got it from. – R.. GitHub STOP HELPING ICE Oct 14 '10 at 13:34
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@AndiDog But sizeof('a') is 2/4 byte...It may b bogus but i didn't get it.. – Parikshita Oct 14 '10 at 13:35
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possible duplicate of [Size of character ('a') in C/C++](http://stackoverflow.com/questions/2172943/size-of-character-a-in-c-c) – Jens Gustedt Oct 14 '10 at 13:38
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@Parixit: See Jens Gustedt's link, taking `sizeof` of a character literal will return an `int`, not a `char`. But that's another topic. – AndiDog Oct 14 '10 at 14:06
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A character literal, such as 'a', will be treated as an integer literal, such as 97 or 0x61. C compilers tend to want every integer to be stored in an int unless told otherwise, so sizeof('a') will probably be sizeof(int).
You should notice, though, that the value of 'a' is less than 127 so it can be stored in a char (which has a maximum value of either 127 or 255 depending on if it is signed or unsigned on your compiler). This is the same as being able to:
unsigned long long x = 0;
unsigned int y = x;
y is assigned from a x whose type is bigger than y's type, but x's value is well within those which y can represent, so no data is lost when the top bits (all 0s) are chopped off.
nategoose
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Along with other people's responces more somewhat related information can be found at this question.
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You've asked a good question. I suggest that you do two things to satisfy yourself that there is enough room to store a character:
1) Look at the documentation for getchar(), and you will see it returns an int, so there is enough room to store a character's value. It is good to understand the underpinnings of the declaration, but if you declare char cTestChar = 'c'; there is enough room to store the value.
2) Generate an assembly listing when you compile. Visual C/C++ supports this as well as gcc. gcc uses the -Wa option to generate assembly output. The map will show the allocation of a character type.
octopusgrabbus
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