As you probably already know, there's no "foreach"-style loop in C.

Although there are already tons of great macros provided here to work around this, maybe you'll find this macro useful:

// "length" is the length of the array.   
#define each(item, array, length) \
(typeof(*(array)) *p = (array), (item) = *p; p < &((array)[length]); p++, (item) = *p)

...which can be used with for (as in for each (...)).

Advantages of this approach:

• item is declared and incremented within the for statement (just like in Python!).
• Seems to work on any 1-dimensional array
• All variables created in macro (p, item), aren't visible outside the scope of the loop (since they're declared in the for loop header).

Disadvantages:

• Doesn't work for multi-dimensional arrays
• Relies on typeof(), which is a GNU extension, not part of standard C
• Since it declares variables in the for loop header, it only works in C11 or later.

Just to save you some time, here's how you could test it:

typedef struct {
    double x;
    double y;
} Point;

int main(void) {
    double some_nums[] = {4.2, 4.32, -9.9, 7.0};
    for each (element, some_nums, 4)
        printf("element = %lf\n", element);

    int numbers[] = {4, 2, 99, -3, 54};
    // Just demonstrating it can be used like a normal for loop
    for each (number, numbers, 5) { 
        printf("number = %d\n", number);
        if (number % 2 == 0)
                printf("%d is even.\n", number);
    }

    char* dictionary[] = {"Hello", "World"};
    for each (word, dictionary, 2)
        printf("word = '%s'\n", word);

    Point points[] = {{3.4, 4.2}, {9.9, 6.7}, {-9.8, 7.0}};
    for each (point, points, 3)
        printf("point = (%lf, %lf)\n", point.x, point.y);

    /* Neither p, element, number or word are visible outside the scope of
    their respective for loops. Try to see if these printfs work (they shouldn't): */
    //printf("*p = %s", *p);
    //printf("word = %s", word);

    return 0;
}

Seems to work on gcc and clang by default; haven't tested other compilers.