Question

With this code, I get a segmentation fault:

   char* inputStr = "abcde";
   *(inputStr+1)='f';

If the code was:

   const char* inputStr = "abcde";
   *(inputStr+1)='f';

I will get compile error for “assigning read-only location”.
However, for the first case, there is no compile error; just the segmentation fault when the assign operation actually happened.

Can anyone explain this?

Answer 1

Here is what the standard says about string literals in section [2.13.4/2]:

A string literal that does not begin with u, U, or L is an ordinary string literal, also referred to as a narrow string literal. An ordinary string literal has type “array of n const char”, where n is the size of the string as defined below; it has static storage duration (3.7) and is initialized with the given characters.

So, strictly speaking, “abcde” has type

const char[6]

Now what happens in your code is an implicit cast to

char*

so that the assignment is allowed. The reason why it is so is, likely, compatibility with C. Have a look also at the discussion here: http://learningcppisfun.blogspot.com/2009/07/string-literals-in-c.html

Once the cast is done, you are syntactically free to modify the literal, but it fails because the compiler stores the literal in a non writable segment of memory, as the standard itself allow.