Can someone tell me why malloc wont work in this case?
(note, line 13 is the char line)
My error is:
1>c:\documents and settings\user\my documents\visual studio 2008\projects\code\code\code.c(13) : error C2143: syntax error : missing ';' before 'type'Code:char * ptr; int main (int argc, char *argv[]) { ptr = malloc(9000); char cmd[50] = ""; return 0; }
Last edited by someprogr; 12-27-2007 at 06:53 PM.
You need to provide a small but complete example of code that illustrates your problem. If there is a line 13 in your sample, that means you have removed other code which might be contributing to the problem. When an error message about a missing semi-colon occurs, the real cause can be code several lines before the compiler gets confused enough to complain.
Assuming your code is C (not C++), and that you are not using a C++ compiler with C code, you need to #include <stdlib.h>
The compiler gives you the line where it concluded that there is an error, not where the error actually is. Errors like these usually spring from something not being quite right on the previous line. Perhaps you're missing a semicolon on line 12?
ptr is global, therefore the initialization of ptr counts as code, not declaration. You can't have declarations following code. You must call malloc() after declaring char cmd[]
Accommodating to a lower denominator is good practice, for portability reasons, since C89 forbids the mixing of declaration and code. However is fine if you are compiling with C99.Originally Posted by brewbuck
Best just put that char* ptr into main instead.
