syntax error before "char" in C

Question

I have the following piece of code, which when I compile it, I get:

smash.c:22 error: syntax error before "char"

I don't understand where the problem is. (line 22 is marked by /*22*/ in the error message, but line numbers do not appear in the code). How can I correct this error?

/*some comments...*/
/*some more comments...*/

#include <stdio.h>
#include <string.h>
#include <unistd.h>
#include <stdlib.h>
#include <signal.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <errno.h>

#include "dir_handling.h"
#include "var_handling.h"



#define     MAXLENGTH           80        

void error_print (char* str)
{
    /*22*/ char *error_message=(*char)malloc((strlen(str)+strlen("smash error: > \"\"\n")+1)*sizeof(char));
    strcpy (error_message,"smash error: > \"");
    strcat(error_message,str);
    strcat(error_message,"\"\n");
    perror (error_message);
    free (error_message);
    // printf ("smash error: > \"%s\"\n",str);
}
...

score 3 · Answer 1 · answered Apr 10 '11 at 23:40

3

(*char) should be (char *).

answered Apr 10 '11 at 23:40

Andrey Vlasovskikh

16,489
7
44
62

score 3 · Accepted Answer · answered Apr 10 '11 at 23:41

3

It should be (char*), as in:

char *error_message=(char*)malloc( etc...

But note: it is good practice not to cast the return of malloc...

answered Apr 10 '11 at 23:41

Mitch Wheat

295,962
43
465
541

If I don't cast the return of `malloc` wouldn't it make problems later? – SIMEL Apr 10 '11 at 23:49
@Ilya Melamed: no, it would not. – Mitch Wheat Apr 10 '11 at 23:50
4

@Ilya Melamed: Quite the opposite, *casting* the result of `malloc` can potentially lead to problems later. – AnT stands with Russia Apr 10 '11 at 23:56
Assuming that I don't forget to include `` are there any other problems, because it does makes the code more readable (in my opinon)? – SIMEL Apr 11 '11 at 00:10

syntax error before "char" in C

2 Answers2