6.8b — C-style string symbolic constants

C-style string symbolic constants

In the lesson 6.6 -- C-style strings, we discussed how you could create and initialize a C-style string, like this:

C++ also supports a way to create C-style string symbolic constants using pointers:

While these above two programs operate and produce the same results, C++ deals with the memory allocation for these slightly differently.

In the fixed array case, the program allocates memory for a fixed array of length 5, and initializes that memory with the string “Alex\0”. Because memory has been specifically allocated for the array, you’re free to alter the contents of the array. The array itself is treated as a normal local variable, so when the array goes out of scope, the memory used by the array is freed up for other uses.

In the symbolic constant case, how the compiler handles this is implementation defined. What usually happens is that the compiler places the string “Alex\0” into read-only memory somewhere, and then sets the pointer to point to it. Multiple string literals with the same content may point to the same location. Because this memory may be read-only, and because making a change to a string literal may impact other uses of that literal, best practice is to make sure the string is const. Also, because strings declared this way are persisted throughout the life of the program (they have static duration rather than automatic duration like most other locally defined literals), we don’t have to worry about scoping issues. Thus, the following is okay:

In the above code, getName() will return a pointer to C-style string “Alex”. This is okay since “Alex” will not go out of scope when getName() terminates, so the caller can still successfully access it.

Rule: Feel free to use C-style string symbolic constants if you need read-only strings in your program, but always make them const!

std::cout and char pointers

At this point, you may have noticed something interesting about the way std::cout handles pointers of different types.

Consider the following example:

On the author’s machine, this printed:


Why did the int array print an address, but the character arrays printed strings?

The answer is that std::cout makes some assumptions about your intent. If you pass it a non-char pointer, it will simply print the contents of that pointer (the address that the pointer is holding). However, if you pass it an object of type char* or const char*, it will assume you’re intending to print a string. Consequently, instead of printing the pointer’s value, it will print the string being pointed to instead!

While this is great 99% of the time, it can lead to unexpected results. Consider the following case:

In this case, the programmer is intending to print the address of variable c. However, &c has type char*, so std::cout tries to print this as a string! On the author’s machine, this printed:


Why did it do this? Well, it assumed &c (which has type char*) was a string. So it printed the ‘Q’, and then kept going. Next in memory was a bunch of garbage. Eventually, it ran into some memory holding a 0 value, which it interpreted as a null terminator, so it stopped. What you see may be different depending on what’s in memory after variable c.

This case is somewhat unlikely to occur in real-life (as you’re not likely to actually want to print memory addresses), but it is illustrative of how things work under the hood, and how programs can inadvertently go off the rails.

6.9 -- Dynamic memory allocation with new and delete
6.8a -- Pointer arithmetic and array indexing

80 comments to 6.8b — C-style string symbolic constants

  • Astronoid

    Hey Alex,
    Thank you for your efforts; I am confused about many points in that lesson.

    The first one is why the compiler treat "&c" as a string not a pointer; the second one is how to get the memory address of that pointer. And thanks;

    • Alex

      std::cout was designed to treat objects of type (const) char* as strings since this is the native type of C-style string literals. Most of the time, this is what we want. Occasionally it causes issues.
      I’m not sure what you actually mean by “get the memory address of that pointer”. &c returns a pointer to variable c, but it doesn’t have an address itself since it’s an r-value.

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