Issue
I'm trying to run a C program in Ubuntu (using the gcc compiler), and for some reason it's not allowing me to use the strcpy function. On the second line of code below:
char test[10];
strcpy(test, "Hello!");
char c[2] = "A";
strcpy(test, c);
I get the following errors:
testChTh.c:56:14: error: expected ‘)’ before string constant
strcpy(test, "Hello!");
^
testChTh.c:59:1: warning: data definition has no type or storage class
strcpy(test, c);
^
testChTh.c:59:1: warning: type defaults to ‘int’ in declaration of ‘strcpy’ [-Wimplicit-int]
testChTh.c:59:1: warning: parameter names (without types) in function declaration
testChTh.c:59:1: error: conflicting types for ‘strcpy’
In file included from testChTh.c:3:0:
/usr/include/string.h:125:14: note: previous declaration of ‘strcpy’ was here
extern char *strcpy (char *__restrict __dest, const char *__restrict __src)
I've included the following headers:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
I've tried using strcpy in a new file with nothing extra, with the same error. I've also tried using:
memset(test, '\0', sizeof(test));
immediately before using strcpy, to no avail.
I've checked all of my opening parenthesis, and they all have a corresponding closing ). Also, when I comment out the strcpy line, the error goes away.
Any insight is much appreciated.
Solution
char test[10];
strcpy(test, "Hello!");
char c[2] = "A";
strcpy(test, c);
If I understand correctly, you have those lines lines at file scope. The line
strcpy(test, "Hello!");
is a statement, and statements are legal only inside a function body. Because the compiler wasn't expecting a statement at that point, it tried to interpret that line as a declaration.
The following, based on your code, is legal (though it doesn't do anything useful):
#include <string.h>
int main(void) {
char test[10];
strcpy(test, "Hello!");
char c[2] = "A";
strcpy(test, c);
}
Answered By - Keith Thompson Answer Checked By - Timothy Miller (WPSolving Admin)