C array of struct pointers - assigning new values

Question

What am I missing? I'm a beginner in C, but still the result I get is unexpected for me. I simplified the case to get only the error.

struct str {
 int a;
};
struct str* strArray[100];
int main() {
 for (int i = 0; i < 100; i++) {
 struct str str1;
 str1.a=i;
 strArray[i]=&str1;
 }
 for(int i = 0; i < 100; i++) {
 printf("i:\t%d\n",strArray[i]->a);
 }
 return 0;
}

The result printed is a hundred 99s. What I would expect is: 1, 2, 3 ...

• Ed Heal: Heard of scope? Things disappear when they go out of scope
• neuromouse: I see ... Sorry, I'm used to programming in Java, things behave differently there ...
• The Paramagnetic Croissant: @neuromouse if you are programming in a different language, you shouldn't expect the same behavior.
• Jim Balter: " What I would expect is: 1, 2, 3 ..." -- Really? Where is the memory where all those different values reside? You have defined only one struct str, namely str1. strArray only contains pointers, and they all point to str1. In fact, your program has undefined behavior because you're accessing str1 outside its scope, so the program could print anything at all (or nothing).
• Jim Balter: " I'm a beginner in C, but still the result I get is unexpected for me." -- That's an odd way to put it. Results are often unexpected by beginners. It would make far more sense if someone said "I'm an old hand at C, but still the result I get is unexpected for me".

Answer 1

This declaration inside the loop

struct str str1;

produces a variable of type struct str that is valid only for the duration of the iteration of the loop in which it has been created. Once the iteration is over, str1 becomes invalid. If you decide to access it by pointer outside the scope of that iteration, you get undefined behavior.

You can use malloc to make "durable" structs for use later in the program:

strArray[i]= malloc(sizeof(struct str));
strArray[i]->a = i;

Since your structs are allocated dynamically, you need to free them once you are done using strArray elements:

for(int i = 0; i < 100; i++) {
 free(strArray[i]);
}

Answer 2

Whole strArray is pointing to str1 which contains 99 at last so your whole array locations pointing to 99 which str1 contains at last of first loop iteration.

• Jim Balter: The value of str1.a is undefined after the first loop completes (in fact, after each iteration of that loop completes) ... it need not be 99.