What happens if you put ';' after for(...)?

If you can use C99 then declaring the loop counter within the for statement might avoid some of these ; issues:

Code:

---

#include <stdio.h>

int main(void)
{
    int array[10] = {0};
    for (int i = 0; i < 10; ++i);
    {
        printf("&#37;d\n", array[i]);
    }
    return 0;
}

---

Quote:

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8 C:\file.c `i' undeclared (first use in this function)

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since i won't be in scope after the loop.

Quote:

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My bloodshed i have a feeling that doesnt shows warnings, just errors.

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If you are using Dev-C++ with Mingw then you should be able to see warnings by typing this in Compiler settings: -Wall -Wextra (look up references for more specific warnings) and -std=c99 to compile in C99 mode (not C90 which doesn't allow declaring the loop counter there).

> So thats no a symbol to decalarate end of text but to declarate placing of a variable there wright?
Correct sir! Basically they're called "specifiers", but they have many names :)

Depending on the "specifier" will depend on how the argument is treated, so each specifier is replaced with an argument. Such that, specifier 1 gets replaced with argument 2 (that's the first argument AFTER the format string). etc.

Eg:
printf(FORMAT STRING, ARG 2, ARG 3, ..., ARGn)

See http://www.cplusplus.com/reference/c...io/printf.html for a list of specifiers and their modifiers.