Why my program crashes when I declare this two variables? - c

Question

I'm trying to make an interpreter using C programming language, it works well, but when I try to add these two variables in the code, the program just crashes.
Fortunately, when I add the static keyword or I leave the variables as global in the program, the program doesn't crash.
Why does this happen ?
Here's the code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <windows.h>
//If I leave the variables here the program doesn't crash
//int esMain = 0;
//int esEnd = 0;
int main(int argc, char** argv){
FILE *f;
f = fopen(argv[1], "r+");
if(f == NULL){
perror("Error: No se encuentra el archivo\nDescripcion");
exit(1);
}
if(ferror(f)){
printf("Archivo corrupto");
exit(1);
}
printf("\nEjecutando archivo: %s\n\n\n", argv[1]);
int esMain = 0;//These two variables makes the program to crash
int esEnd = 0;
//But if I add the static keyword in both variables the program works well.
char* str;
while(1){
fgets(str, 25, f);
if(strncmp(str, "MAIN:", 5) == 0){
esMain = 1;
}
if(strncmp(str, "END.", 4) == 0){
esEnd = 1;
}
if(feof(f) != 0){
if(esMain == 0){
printf("No existe el parametro main, cierre por no tener el parametro main (poner MAIN: ENCIMA)");
exit(1);
}
if(esEnd == 0){
printf("No existe el parametro end, cierre por no tener el parametro main (poner END. ENCIMA)");
exit(1);
}
break;
}
}
rewind(f);
while(1){
fgets(str, 500, f);
if(strncmp(str, "print ", 6) == 0){
printf(str + 6);
}
if(strncmp(str, "msg", 3) == 0){
if(strstr(str + 4, "check")){
MessageBox(HWND_DESKTOP, str + 10, "Check", MB_ICONINFORMATION);
}
if(strstr(str + 4, "error")){
MessageBox(HWND_DESKTOP, str + 10, "Error", MB_ICONERROR);
}
}
if(strncmp(str, "pause", 5) == 0){
getch();
}
if(feof(f) != 0){
break;
}
}
printf("\n\n\nEND EXECUTION");
fclose(f);
return 0;
}

Answer 1

The char* str; declaration is probably what is breaking your code. When you declare it as global it is stored in a different place in memory than if you declare it inside a function.
You got lucky to have your program working with it as global variable. Since you haven't reserved any place in memory, the variable is accessing some memory that it shouldn't (undefined behavior). Lucky might not be the best word to describe because due to t his undefined behavior you might think that your program is working correctly, which is not (If it had crashed you would be 100% that there were errors).
What you can do to fix that is one of the following:
allocate it dynamically: char *str = (char*)malloc(sizeof(char)*25);
change it to br an array: char str[25];
point to an existing array: char arr[25]; char *str = arr;