strcpy() causes segmentation fault? [duplicate]
Possible Duplicate:
Getting Segmentation Fault
Why does this code cause a segmentation fault?
char *text = "foo";
strcpy(text, "");
As far as I understand it, the first line allocates some memory (to hold the string "foo") and text
points to that allocated memory. The second line copies an empty string into the location that text
points to.
This code might not make a lot of sense, but why does it fail?
Whenever you have a string literal (in your case, "foo"), the program stores that value in a readonly section of memory.
strcpy
wants to modify that value but it is readonly, hence the segmentation fault.
Also, text
should be a const char*
, not a char*
.
Because a string literal (like "foo"
) is read-only.
Because the string literals are stored in the read only region of memory.
Thus attempting the modification of foo
(using strcpy
in this case) is an undefined behavior.
精彩评论