Assign a string literal to a char*

Question

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How to get rid of deprecated conversion from string constant to ‘char*’ warnings in GCC?

This assignment:

char *pc1 = "test string";

gives me this warning:

warning: deprecated conversion from string constant to 'char*'

while this one seems to be fine:

char *pc2 = (char*)("test string");

Is this one a really better way to proceed?

Notes: for other reasons I cannot use a const char*.

casting away the warning is really a bad idea. Just use `char const*pc1` and everything should be fine. — Jens Gustedt, Dec 02 '11 at 12:41
You might not want to use `const char*`, but that string literal really is a `const`. You can't modify it. — David Heffernan, Dec 02 '11 at 12:50
As it's C++ code, you could have tagged the question with c++ ... and use `std::string` instead — pmg, Dec 02 '11 at 15:01
@OliCharlesworth, MarceloCantos, DavidHeffernan - I cannot: I must pass it to a function that needs a char*. And I cannot modify that function. — Pietro M, Dec 02 '11 at 15:02
@pmg You are right, but this is C++ code that calls C functions. So the question is related to the C part of C++. — Pietro M, Dec 02 '11 at 15:05

score 11 · Answer 1 · answered Dec 02 '11 at 12:25

11

A string literal is a const char[] in C++, and may be stored in read-only memory so your program will crash if you try to modify it. Pointing a non-const pointer at it is a bad idea.

answered Dec 02 '11 at 12:25

Wyzard

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Whether or not it is stored in read-only memory, it's UB to modify it. End of story. – David Heffernan Dec 02 '11 at 14:12

NPE · Answer 2 · 2011-12-02T12:49:29.800

7

In your second example, you must make sure that you don't attempt to modify the the string pointed to by pc2.

If you do need to modify the string, there are several alternatives:

Make a dynamically-allocated copy of the literal (don't forget to free() it when done):

char *pc3 = strdup("test string"); /* or malloc() + strcpy() */
Use an array instead of a pointer:

char pc4[] = "test string";

edited Dec 02 '11 at 12:49

answered Dec 02 '11 at 12:20

NPE

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+1: Although you should point out that `strdup` is not standard C... – Oliver Charlesworth Dec 02 '11 at 12:25
I used `char pc4[] = "test string";` and I get no warnings. I wanted to know if this is the correct way to proceed, or if I am just planting a bug in my code... – Pietro M Dec 02 '11 at 15:24
@PietroM: No, you're not planting any bugs, as long as you bear in mind that the allocated array is `strlen("test string")+1` bytes long. If you need a longer array, you can specify the length explicitly: `char pc4[32] = "test string";` – NPE Dec 02 '11 at 15:28
To be more precise, I used `static char pc4[] = "test string";` and I get no warnings. I did it static because it is inside a function, and I want the string to be available even when out of the function. I wanted to know if this is the correct way to proceed, or if I am just planting a bug in my code... – Pietro M Dec 02 '11 at 15:33

score 5 · Answer 3 · answered Dec 02 '11 at 12:20

5

That depends on whether you need to modify the string literal or not. If yes,

char pc1[] = "test string";

answered Dec 02 '11 at 12:20

Mahesh

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That doesn't let you modify the literal, it let's you modify a copy – David Heffernan Dec 02 '11 at 13:10
Correct. Your comment is precise. – Mahesh Dec 02 '11 at 13:38

Assign a string literal to a char*

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