问题
Yesterday I had a little trouble with a homemade "strcpy" function. It works now though but I'm a little confused!
char* a = "Hello, World!"; //Works
char b[] = "Hello, World!"; //Works also
strcpy(a, "Hello!"); //Segmentation fault
strcpy(b, "Haha!!"); //Works..
Where is the difference? Why does the char pointer cause a "Segmentation fault"?
Why does this even work? :
char* a = "Haha"; //works
a = "LOL"; //works..
回答1:
char* a = "Hello, World!";
gives you a pointer to a string literal. A string literal may exist in read only memory so its contents cannot be changed.
char* a = "Haha"; //works
a = "LOL"; //works..
changes the pointer a to point to a different string literal. It doesn't attempt to modify the contents of either string literal so is safe/correct.
char b[] = "Hello, World!"
declares an array on the stack and initialises it with the contents of a string literal. Stack memory is writeable so its perfectly safe to change the contents of this memory.
回答2:
In your first example since you are trying to write to a read only memory pointed by a,you will get the segmentation fault.If you want to use pointers then allocate the memory on heap,use and delete it after its use. Where as b is an array of chars initialized with "Hello, World!".
In the second example you are making the pointer to point to different string literal which should be fine.
来源：https://stackoverflow.com/questions/16692887/difference-between-array-and-pointer