Pass struct by reference in C

Question

Is this code correct? It runs as expected, but is this code correctly using the pointers and dot notation for the struct?

struct someStruct {  unsigned int total; };  int test(struct someStruct* state) {  state->total = 4; }  int main () {  struct someStruct s;  s.total = 5;  test(&s);  printf("\ns.total = %d\n", s.total); } 

Answer 1

That's correct usage of the struct. There are questions about your return values.

Also, because you are printfing a unsigned int, you should use %u instead of %d.

Answer 2

Your use of pointer and dot notation is good. The compiler should give you errors and/or warnings if there was a problem.

Here is a copy of your code with some additional notes and things to think about so far as the use of structs and pointers and functions and scope of variables.

Note: A code writing difference in the source example below is I put a space after the struct name and before the asterisk in the function definition/declaration as in struct someStruct *p1; and the OP put a space after the asterisk as in struct someStruct* p1;. There is no difference to the compiler, just a readability and habit difference for the programmer. I prefer putting the asterisk next to the variable name to make clear the asterisk changes the variable name it is next to. This is especially important if I have more than one variable in a declaration or definition. Writing struct someStruct *p1, *p2, var1; will create two pointers, p1 and p2, and a variable, var1. Writing struct someStruct* p1, p2, var1; will create single pointer, p1 and two variables p2 and var1

// Define the new variable type which is a struct. // This definition must be visible to any function that is accessing the // members of a variable of this type. struct someStruct {     unsigned int total; };  /*  * Modifies the struct that exists in the calling function.  *   Function test() takes a pointer to a struct someStruct variable  *   so that any modifications to the variable made in the function test()  *   will be to the variable pointed to.  *   A pointer contains the address of a variable and is not the variable iteself.  *   This allows the function test() to modify the variable provided by the  *   caller of test() rather than a local copy.  */ int test(struct someStruct *state) {     state->total = 4;     return 0; }  /*   * Modifies the local copy of the struct, the original  * in the calling function is not modified.  * The C compiler will make a copy of the variable provided by the  * caller of function test2() and so any changes that test2() makes  * to the argument will be discarded since test2() is working with a  * copy of the caller's variable and not the actual variable.  */ int test2(struct someStruct state) {     state.total = 8;     return 0; }  int test3(struct someStruct *state) {     struct someStruct  stateCopy;     stateCopy = *state;    // make a local copy of the struct     stateCopy.total = 12;  // modify the local copy of the struct     *state = stateCopy;    /* assign the local copy back to the original in the                               calling function. Assigning by dereferencing pointer. */     return 0; }  int main () {     struct someStruct s;      /* Set the value then call a function that will change the value. */     s.total = 5;     test(&s);     printf("after test(): s.total = %d\n", s.total);      /*      * Set the value then call a function that will change its local copy       * but not this one.      */     s.total = 5;     test2(s);     printf("after test2(): s.total = %d\n", s.total);      /*       * Call a function that will make a copy, change the copy,        then put the copy into this one.      */     test3(&s);     printf("after test3(): s.total = %d\n", s.total);      return 0; }