assert() with message

Question

I saw somewhere assert used with a message in the following way:

assert(("message", condition));

This seems to work great, except that gcc throws the following warning:

warning: left-hand operand of comma expression has no effect

How can I stop the warning?

Answer 1

If you want to pass a formatted message, you could use the following macros:

#include <stdio.h>
#include <errno.h>
#include <string.h>
#include <assert.h>

#define clean_errno() (errno == 0 ? "None" : strerror(errno))
#define log_error(M, ...) fprintf(stderr, "[ERROR] (%s:%d: errno: %s) " M "\n", __FILE__, __LINE__, clean_errno(), ##__VA_ARGS__)
#define assertf(A, M, ...) if(!(A)) {log_error(M, ##__VA_ARGS__); assert(A); }

Then use it like printf:

// With no args
assertf(self != NULL,"[Server] Failed to create server.");

// With formatting args
assertf((self->socket = u_open(self->port)) != -1,"[Server] Failed to bind to port %i:",self->port);
// etc...

Output:

[ERROR] (../src/webserver.c:180: errno: Address already in use) [Server] Failed to bind to port 8080: webserver: ../src/webserver.c:180: server_run: Assertion `(self->socket = u_open(self->port)) != -1' failed.

Based on http://c.learncodethehardway.org/book/ex20.html

Answer 2

You could write your own macro that provides the same usage of _Static_assert(expr, msg):

#include <assert.h>
#include <stdbool.h>
#include <stdio.h>


/*
 * void assert_msg(bool expr, const char *msg);
 */
#if !defined(NDEBUG)
#define assert_msg(expr, msg)   do                  \
{                                                   \
        const bool  e_ = expr;                      \
                                                    \
        if (!e_) {                                  \
                fputs(msg, stderr);                 \
                fputc('\n', stderr);                \
                assert(e_);                         \
        }                                           \
} while (0)
#else
#define assert_msg(expr, msg)   do                  \
{                                                   \
                                                    \
        if (!(expr))                                \
                warn_bug(msg);                      \
} while (0)
#endif

I also have a macro warn_bug() that prints the name of the program, the file, the line, the function, the errno value and string, and a user message, even if asserts are disabled. The reason behind it is that it won't break the program, but it will warn that a bug will probably be present. You could just define assert_msg to be empty if defined(NDEBUG), though.

Answer 3

According to following link http://www.cplusplus.com/reference/clibrary/cassert/assert/

assert is expecting only expression. May be you are using some overloaded function.

According to this, only expression is allowed and thus you are getting this warning.

Answer 4

Use -Wno-unused-value to stop the warning; (the option -Wall includes -Wunused-value).

I think even better is to use another method, like

assert(condition && "message");

Answer 5

Try:

#define assert__(x) for ( ; !(x) ; assert(x) )

use as such:

assert__(x) {
    printf("assertion will fail\n"); 
}

Will execute the block only when assert fails.

IMPORTANT NOTE: This method will evaluate expression x twice, in case x evaluates to false! (First time, when the for loop is checking its condition; second time, when the assert is evaluating the passed expression!)

Answer 6

By tradition, (void) communicates to the compiler that you are knowingly ignoring an expression:

/* picard.c, TNG S6E11. */
#define assertmsg(x, msg) assert(((void) msg, x))
assertmsg(2+2==5, "There! are! four! lights!");