# operator in C

I am reading about the preprocessor features and functions. The author states "If you place a # in front of a parameter in a macro definition, the preprocessor creates a constant string out of the macro argument when the macro is invoked." And one of the examples they give is:

Code:

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#include <stdio.h>
#define printint(var) printf( # var " = %i\n", var)
int main(void)
{
int count = 2;
printint (count);
}

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Now they just say that the #define statement would cause printint to change to printf with the added content. But the above code does not print anything. Do I need anything else I am missing?

No, there is nothing missing. In my own tests, it produces the expected outcome. Maybe your window is closing before you can see the output?

Quote:

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Originally Posted by whiteflags

No, there is nothing missing. In my own tests, it produces the expected outcome. Maybe your window is closing before you can see the output?

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yes; it runs. I am not sure why it did not earlier. I added some more stuff, and now I get error messages. Here is what I have:

Code:

---

#include <stdio.h>
int count, x;
#define printx(n) printint (x ## n)
#define printint(var) printf( # var " = %i\n", var)
int main(void)
{
count = 2;
x = 3;
printx (10);
printint (count);
}

---

Code:

---

$ gcc test.c -o test
test.c: In function 'main':
test.c:3:29: error: 'x10' undeclared (first use in this function)
 #define printx(n) printint (x ## n)
                            ^
test.c:4:48: note: in definition of macro 'printint'
 #define printint(var) printf( # var " = %i\n", var)
                                                ^
test.c:9:1: note: in expansion of macro 'printx'
 printx (10);
 ^
test.c:3:29: note: each undeclared identifier is reported only once for each function it appears in
 #define printx(n) printint (x ## n)
                            ^
test.c:4:48: note: in definition of macro 'printint'
 #define printint(var) printf( # var " = %i\n", var)
                                                ^
test.c:9:1: note: in expansion of macro 'printx'
 printx (10);
 ^

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I tried declaring the variables globally. I am not sure what else to do.

Did you compile it?

Calling the program 'test' is a bad idea in Unix/Linux, and other environments which use a unix-like shell (such as mingw).
test is a built-in command of many shells, so just typing 'test' invokes that built-in.

You might get further if you type in ./test.exe
assuming that the program is called test.exe

Quote:

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Originally Posted by Salem

Did you compile it?

Calling the program 'test' is a bad idea in Unix/Linux, and other environments which use a unix-like shell (such as mingw).
test is a built-in command of many shells, so just typing 'test' invokes that built-in.

You might get further if you type in ./test.exe
assuming that the program is called test.exe

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Thanks. It worked. I moved the content to another file. Here is the corrected file:

Code:

---

#include <stdio.h>
int count, x10=1;
#define printx(n) printint (x ## n)
#define printint(var) printf( # var " = %i\n", var)
int main(void)
{
count = 2;
printx (10);
printint (count);
}

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