Question

#include <stdio.h>

typedef struct point{
    int x; 
    int y;
};

void main (void){

    struct point pt;
    pt.x = 20;
    pt.y = 333;

    struct point pt2;
    pt2.y = 55;

    printf("asd");
    return;
}

VS 2008

c:\documents and settings\lyd\mis documentos\ejercicio1.c\ejercicio1.c\ejercicio1.c(14) : error C2143: syntax error : missing ';' before 'type'
c:\documents and settings\lyd\mis documentos\ejercicio1.c\ejercicio1.c\ejercicio1.c(15) : error C2065: 'pt2' : undeclared identifier
c:\documents and settings\lyd\mis documentos\ejercicio1.c\ejercicio1.c\ejercicio1.c(15) : error C2224: left of '.y' must have struct/union type
Build log was saved at "file://c:\Documents and Settings\LYD\Mis documentos\ejercicio1.c\ejercicio1.c\Debug\BuildLog.htm"
ejercicio1.c - 3 error(s), 0 warning(s)
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

Answer 1

Since the question is tagged C (and not C++), and since the compiler is MSVC 2008, you are stuck with C89 semantics. That means you cannot declare variables in a block after the first statement. Hence, the second struct variable is not allowed there. (Both C99 and C++ allow you declare variables at any point in the block. Go tell MS to update their C compiler to support C99.)

Your other bug is that main() returns an int, hence:

#include <stdio.h>

struct point
{
    int x; 
    int y;
};

int main (void)
{
    struct point pt;
    struct point pt2;
    pt.x = 20;
    pt.y = 333;
    pt2.x = 4;
    pt2.y = 55;
    printf("asd");
    return 0;
}

---

Some hours later: the keyword typedef is not needed in the code because no name is specified after the close brace and before the semi-colon. This doesn’t stop it compiling; it will elicit a warning with the compiler set fussy.