Assigning multiple characters in an int in C language

The character type data is stored by its ASCII value internally in C. If we want to print a single character as integer, we will get the ASCII value. But when we are trying to print more than one character using a single quote, then it will print some strange output.

Syntax

int variable = 'character';
int variable = 'multiple_chars';  // Non-standard behavior

Example

Please check the following program to get the idea −

#include <stdio.h>

int main() {
    printf("%d
", 'A');
    printf("%d
", 'AA');
    printf("%d
", 'ABC');
    return 0;
}

Output

65
16705
4276803

How It Works

The ASCII of A is 65. So at first it is showing 65 (binary: 01000001). Now for AA, it is showing 16705. This is formed by packing two ASCII values together:

• 'A' = 65 (0x41)
• 'AA' = 0x4141 = 16705 in decimal
• 'ABC' = 0x414243 = 4276803 in decimal

Each character is stored as one byte, and multiple characters are packed into the integer using little-endian or big-endian byte order depending on the system architecture.

Key Points

• Using multiple characters in single quotes is non-standard behavior and should be avoided.
• The result depends on the compiler and system architecture.
• For multiple characters, use string literals with double quotes instead.

Conclusion

While C allows multiple characters in single quotes, this creates implementation-defined behavior. It's better to use proper string handling with double quotes for multiple characters to ensure portability and clarity.