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I am new in programming Language and I need your help here. I am here studying someone's code and I can across with these expressions, my doubt is how is the operation done here, given that a character and an integers are two different data types? How will the integer type hold the character value?

Thanks

int line, col;
char ch;
scanf("%d%c", &line, &ch);
//line--;
col = ch - 'A';
Osmany Jr
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    because of [promotions](https://stackoverflow.com/a/46073296/3476780) – yano Feb 01 '22 at 22:00
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    Note too that `'A'` is type `int` (*not* `char`). The `ch` is promoted to `int`, then `'A'` is subtracted, and that is assigned to `col`, which is a large enough integer type to hold the result. – Weather Vane Feb 01 '22 at 22:05

3 Answers3

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my doubt is how is the operation done here, given that a character and an integers are two different data types?

I'm unsure how well this question will be received here, given it being about some fairly basic behavior of the language, but I commend you for thinking about type and type matching. Keep doing that!

The first thing to understand is that in C, char and its signed and unsigned variants are among the integer data types, so there is no mismatch of type category, just a question of possibly-different range and signedness. Characters are represented by integer codes (as, indeed, is pretty much everything in the computer's memory).

The second thing to understand is that C supports all manner of arithmetic operations on operands of mixed types. It defines a set of "usual arithmetic conversions" that are used to choose a common type for the operands and the result of each arithmetic operation. The operands are automatically converted to that type. I won't cover all the details here, but basically, floating-point types win over integer types, and wider types win over narrower types.

The third thing to understand is that C does not in any case directly define arithmetic on integer types narrower than (technically, having integer conversion rank less than that of) int. When a narrower value appears in an arithmetic expression, it is automatically converted to int (if int can represent all values of the original type) or to unsigned int. These automatic conversions are called the "integer promotions", and they are a subset of the usual arithmetic conversions.

A fourth thing that is sometimes important to know is that in C "integer character constants" such as 'A' have type int, not type char (C++ differs here).

So, to evaluate this ...

col = ch - 'A';

... the usual arithmetic conversions are first applied to ch and 'A'. This involves performing the integer promotions on the value of ch, resulting in the same numeric value, but as an int. The constant 'A' already has type int, so these now match, and their difference can be computed without any further conversions. The result is an int, which is the same type as col, so no conversion is required to assign the result, either.

How will the integer type hold the character value?

Character values are integer values. Type int can accommodate all values that type char can accommodate.* Nothing special is happening in that regard.

*Technically, int can accommodate all values that can be represented by signed char, unsigned int can accommodate all values that can be represented by type unsigned char, and at least one of the two can accommodate all values that can be represented by (default) char. You are fairly unlikely to run across a C implementation where there are char values that int cannot accommodate, and the above assumes that you are not working with such an implementation, but these are allowed and some may exist.

John Bollinger
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At the fundamental level, every type in C (be it char, int, uint32_t, short, long...) is represented by bytes, and is 'numerical' in form. You can subtract them from each other / add them together in whichever combination you like - as long as you store the resulting value in a variable of a type which is big enough to hold it - otherwise it will cause a buffer overflow.

In your example, since a char type is represented by a single byte, and an int is composed of 8, the result of this subtraction will simply be stored in the right-most byte of an int (however, depending on if you're dealing with an expression which will yield a negative value, then the representation of the int in memory will be slightly different - look into 2's complement if you're interested).

alka04
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  • `int` is typically four bytes in current C implementations, not eight, and the C standard allows it to vary. When the result of arithmetic is out of range of the result type, that is just “overflow,” not “buffer overflow.” “Buffer overflow” refers to accessing elements outside the range of an array. And storing a value in an integer type that cannot represent it is not what the C standard calls overflow. For an unsigned type, the value is wrapped modulo 2^w, where w is the width of the type. For a signed type, an implementation-defined value or signal is produced. – Eric Postpischil Feb 01 '22 at 23:46
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When you subtract two characters and put them in a variable of integer type, in fact the ASCII code of the two characters is subtracted. For example when you have:

int col = 'D' - 'A';

The value of col is equal to 3 Because ascii code of D is equal to 68 and ascii code of A is 65. So col is 3, however D & A were character. Also you can see here

Ali
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  • The C standard does not require implementations to use ASCII. – Eric Postpischil Feb 01 '22 at 22:57
  • Nor does it require that the integer values assigned to the alphabet characters be consecutive, so there is no guarantee that `'D' - 'A'` will be `3`. It does, however, require that the 10 decimal digit characters be ordered and consecutive, so that `'3' - '0'` will always be `3`. – Weather Vane Feb 01 '22 at 23:40
  • Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Feb 02 '22 at 08:01