4 ms·

In the last example, > char string[] = "Hello world"; I thought that gives out warning these days. Isn't the proper type of a string literal the following? >

by wbkang 10y ago

In the last example, > char string[] = "Hello world"; I thought that gives out warning these days. Isn't the proper type of a string literal the following? > const char string[] = "Hello world"; Therefore, you can't really modify individual characters there.

smitherfield 10y ago
C string literals are " char * ". C++ string literals are " const char * " (one of the minor incompatibilities between the languages). They ought to be "const char[]" but aren't. And, the behavior of "char[]" should be the same as "struct { char[] }", but it isn't.
- gpderetta 10y ago
  They actually are proper arrays in C++ [1], although, as usual, they decay to pointers faster than an unstable particle. [1]: https://godbolt.org/g/ZWISCu https://godbolt.org/g/ZWISCu
  - smitherfield 10y ago
    I stand corrected!
kbp 10y ago
You're confusing things a little bit. 'char s[] = "foo"' is equivalent to 'char s[] = { 'f', 'o', 'o', '\0' }' and gives you a normal mutable array (or you can make it const and then it's const; no surprises); it is okay to mutate that array. 'char * s = "foo"' gives you a pointer to a string literal that might be placed in read-only storage; in C++ it is a const char * (so modern compilers should complain about that initialisation), while in C it is a char * where it's UB if you modify it (compilers may warn about storing it in a non-const char *; many don't).
coldpie 10y ago
> Isn't the proper type of a string literal the following? No, it's a const char pointer. Your declaration actually makes a copy of the "Hello world" string into a new char array, distinct from the string literal itself: [~]$ cat test.c #include <stdio.h> int main(int argc, char ** argv) { char a[] = "test2"; printf("sizeof(a): %lu\n", sizeof(a)); return 0; } [~]$ gcc -std=c99 -pedantic -Wall -o test test.c [~]$ ./test sizeof(a): 6 [~]$
- copascetic 10y ago
  The type of a string literal is array of char, and like other arrays, they decay to pointers to the first element when used in expression context, with three exceptions: sizeof, taking the address with &, and when used as a string initializer. "test2" in your example is not a const char *, it's an initializer, since this is one of the exceptions to array-to-pointer decay. http://c-faq.com/aryptr/aryptrequiv.html http://c-faq.com/aryptr/aryptrequiv.html http://c-faq.com/decl/strlitinit.html http://c-faq.com/decl/strlitinit.html

by wbkang 10y ago

smitherfield 10y ago

C string literals are " char * ". C++ string literals are " const char * " (one of the minor incompatibilities between the languages). They ought to be "const char[]" but aren't. And, the behavior of "char[]" should be the same as "struct { char[] }", but it isn't.

gpderetta 10y ago
They actually are proper arrays in C++ [1], although, as usual, they decay to pointers faster than an unstable particle. [1]: https://godbolt.org/g/ZWISCu https://godbolt.org/g/ZWISCu
- smitherfield 10y ago
  I stand corrected!

kbp 10y ago

You're confusing things a little bit. 'char s[] = "foo"' is equivalent to 'char s[] = { 'f', 'o', 'o', '\0' }' and gives you a normal mutable array (or you can make it const and then it's const; no surprises); it is okay to mutate that array. 'char * s = "foo"' gives you a pointer to a string literal that might be placed in read-only storage; in C++ it is a const char * (so modern compilers should complain about that initialisation), while in C it is a char * where it's UB if you modify it (compilers may warn about storing it in a non-const char *; many don't).

coldpie 10y ago

> Isn't the proper type of a string literal the following? No, it's a const char pointer. Your declaration actually makes a copy of the "Hello world" string into a new char array, distinct from the string literal itself: [~]$ cat test.c #include <stdio.h> int main(int argc, char ** argv) { char a[] = "test2"; printf("sizeof(a): %lu\n", sizeof(a)); return 0; } [~]$ gcc -std=c99 -pedantic -Wall -o test test.c [~]$ ./test sizeof(a): 6 [~]$

copascetic 10y ago
The type of a string literal is array of char, and like other arrays, they decay to pointers to the first element when used in expression context, with three exceptions: sizeof, taking the address with &, and when used as a string initializer. "test2" in your example is not a const char *, it's an initializer, since this is one of the exceptions to array-to-pointer decay. http://c-faq.com/aryptr/aryptrequiv.html http://c-faq.com/aryptr/aryptrequiv.html http://c-faq.com/decl/strlitinit.html http://c-faq.com/decl/strlitinit.html