concatenating strings in a printf statement - c

Question

gcc 4.7.2
c89
Hello,
#define LOG_ERR(fmt, ...) \
fprintf(stderr, "[ERROR] %s:%d: error [%s] " fmt "\n", __func__, __LINE__, strerror(errno), ##__VA_ARGS__)
And I am using it like this:
LOG_ERR("Failed to connect to message queue [ %d ]", msg_id);
The fmt has been concatenated in the fprintf statement. How is this possible?
I have tried to do the same with the following below just to test the concept, but failed with a compile error:
/* Using char array */
const char name[] = "Joe";
printf("Hello how " name " how are you today?\n");
Using constant string literal
const char *name = "Joe";
printf("Hello how " name " how are you today?\n");
Both game me the following error:
expected ')' before name
Many thanks for any suggestions,

Answer 1

You need a format specifier, check out the documentation for printf
const char *name = "Joe"; // or your char array, either's fine
printf("Hello how %s how are you today?\n", name);
Your attempt:
printf("Hello how " name " how are you today?\n");
Looks a little more C++ish
cout << "Hello how " << name << "are you today?\n";

Answer 2

When it fails, it's because you're using variables. String literals can be concatenated by the compiler i.e. if you write "abc" "123" then the compiler will treat this as "abc123". And when you do this in the macro, the preprocessor means that this is exactly what is sent to the compiler
LOG_ERR("Failed to connect to message queue [ %d ]", msg_id);
becomes
fprintf(stderr, "[ERROR] %s:%d: error [%s] " "Failed to connect to message queue [ %d ]" "\n", myfunction, 123, strerror(errno), msg_id);
Might be worth checking out the stringizer and concatenation macros too (for the preprocessor - # and ## and I never remember which is which...)
Andrew

Answer 3

Difference is that the macro is textually replacing the word fmt with your string. Putting two or more literal strings together gets you the concatenation of those strings.
"Hello " "World" "!!"
/* is the same as */
"Hello World!!"
Remember, only literal strings will do this. This does not work for variables.
Think of macros like a find/replace in your code. To illustrate, consider your macro definition -
#define LOG_ERR(fmt, ...) \
fprintf(stderr, "[ERROR] %s:%d: error [%s] " fmt "\n",
__func__, __LINE__, strerror(errno), ##__VA_ARGS__)
When you use it like this -
LOG_ERR("Failed to connect to message queue [ %d ]", msg_id);
It does a textual replacement and becomes -
fprintf(stderr, "[ERROR] %s:%d: error [%s] " "Failed to connect to message queue [ %d ]" "\n",
__func__, __LINE__, strerror(errno), msg_id)
The side-by-side strings and concatenated and voila -
fprintf(stderr, "[ERROR] %s:%d: error [%s] Failed to connect to message queue [ %d ]\n",
__func__, __LINE__, strerror(errno), msg_id)

Answer 4

If you want to define a macro for fprintf() you may do like this which is simplest IMHO.
#define LOG_ERR(...) fprintf(stderr, __VA_ARGS__)
and use it like
int main()
{
int myVar =100;
LOG_ERR ("The value of myVar is %d", myVar);
return 0;
}

Answer 5

#include <stdio.h>
#include <sys/time.h>
#include <string>
#define MyPrint(...) {\
struct timeval tv;\
gettimeofday(&tv,0);\
printf("%d.%d:", (unsigned int)tv.tv_sec, (unsigned int)tv.tv_usec);\
printf(__VA_ARGS__);}