Why does strcpy trigger a segmentation fault with global variables?

Question

So I've got some C code: #include #include /* putting one of the "char*"s here causes a segfault */ void main() { char* path = "/temp"; char* temp; strcpy(temp, path); } This compiles, runs, and behaves as it looks. However, if one or both of the character pointers is declared as global variable, strcpy results in a segmentation fault. Why does this happen? Evidently there's an error in my understanding of scope. JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

Answer 1

As other posters mentioned, the root of the problem is that temp is uninitialized. When declared as an automatic variable on the stack it will contain whatever garbage happens to be in that memory location. Apparently for the compiler+CPU+OS you are running, the garbage at that location is a valid pointer. The strcpy "succeeds" in that it does not segfault, but really it copied a string to some arbitrary location elsewhere in memory. This kind of memory corruption problem strikes fear into the hearts of C programmers everywhere as it is extraordinarily difficult to debug. When you move the temp variable declaration to global scope, it is placed in the BSS section and automatically zeroed. Attempts to dereference *temp then result in a segfault. When you move *path to global scope, then *temp moves up one location on the stack. The garbage at that location is apparently not a valid pointer, and so dereferencing *temp results in a segfault.