## Why Does Concatenating a String Literal and a Character Literal in C Lead to Unexpected Results?

Conundrum: String and Character Concatenation Enigma in C

When attempting to concatenate a string literal with a character literal in C , some unexpected behavior can arise. Consider the following code:

<code class="cpp">string str = "ab" + 'c';
cout << str << endl;</code>

Interestingly, this line results in the output "abc". To unravel this mystery, it's crucial to understand the underlying operations.

Under the hood, the compiler interprets the string literal "ab" as a C-style string and the character literal 'c' as an integer value. Since string concatenation is undefined for these types, the compiler adds the integer value of 'c' to the address of the string literal.

This manipulation can lead to unpredictable outcomes, as the allocated memory for the string is exceeded. Consequently, the program may print characters from the resulting address until encountering a null character.

To resolve this issue, several approaches can be taken. Firstly, you can explicitly convert the character literal to a string using a cast:

<code class="cpp">string str = std::string("ab") + 'c';</code>

Alternatively, you can employ the string concatenation operator, which performs the intuitive concatenation:

<code class="cpp">string str = "ab";
str += 'c';</code>

In the second case, the string class's overloaded operator method is utilized for seamless concatenation.

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