Why Does Concatenating a Character to a String in C Sometimes Lead to Unexpected Output?

Concatenating String and Character Literals in C

In C , the operator can be used for string and character concatenation. However, this behavior varies depending on the type of operands involved.

Consider the following code:

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

This code produces the output "abc" as expected. This is because the string literal "ab" is automatically converted to a const char* pointer. The character literal 'c' is promoted to an integer value and added to the address of the string literal. The result is a new string that contains the characters "abc".

However, consider this alternative:

<code class="cpp">char ch = 'c';
string str1 = "ab";
string str2 = str1 + ch;
cout << str2 << endl;</code>

This code produces the unexpected output "ed".

To understand this behavior, we need to first observe that the string str1 is of type string, not const char*. When we add the character ch to the string, the compiler tries to use the overloaded operator for string objects. However, there is no operator overload for string and char.

To resolve this ambiguity, the compiler promotes ch to an integer and performs the addition as in the first example. However, since the resulting address points past the allocated memory for the string str1, we get undefined behavior. Characters are retrieved from this address until a null character is encountered, resulting in the unexpected output.

To prevent this behavior, we can explicitly cast "ab" to a string before adding 'c':

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

Alternatively, we can use the concatenation operator =:

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

Both of these approaches ensure that the result is a valid string.

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