How Can I Print a Character's Integer Value in C Using `cout`?

Printing Characters as Integers in C Using cout

When attempting to output characters as integers using the cout function, one may encounter unexpected results. This is because cout typically treats characters as characters rather than 8-bit integers. However, character types in C are indeed integral types.

To resolve this issue and ensure that characters are printed as integers, an effective solution is to utilize the unary plus operator ( ). This operator promotes the character to a primitive data type (e.g., int) that can be printed as a number, regardless of its original type.

char a = 0xab;
cout << +a; // Output: 171

By applying the unary plus operator, the character a is cast to an int. This casting preserves the numerical value of the character and allows it to be printed as an integer.

It's important to note that this approach is valid as long as the data type for the character type (char, signed char, or unsigned char) provides a unary plus operator with standard semantics. Moreover, defining a class to represent a number with a unary plus operator that returns the current object's value (by value or by reference-to-const) enables printing the character as a number.

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