C/C++ line number | 易学教程

问题

In the sake of debugging purposes, can I get the line number in C/C++ compilers? (standard way or specific ways for certain compilers)

e.g

if(!Logical)
    printf(\"Not logical value at line number %d \\n\",LineNumber);
    // How to get LineNumber without writing it by my hand?(dynamic compilation)

回答1:

You should use the preprocessor macro __LINE__ and __FILE__. They are predefined macros and part of the C/C++ standard. During preprocessing, they are replaced respectively by a constant string holding an integer representing the current line number and by the current file name.

Others preprocessor variables :

__func__ : function name (this is part of C99, not all C++ compilers support it)
__DATE__ : a string of form "Mmm dd yyyy"
__TIME__ : a string of form "hh:mm:ss"

Your code will be :

if(!Logical)
  printf("Not logical value at line number %d in file %s\n", __LINE__, __FILE__);

回答2:

As part of the C++ standard there exists some pre-defined macros that you can use. Section 16.8 of the C++ standard defines amongst other things, the __LINE__ macro.

__LINE__: The line number of the current source line (a decimal constant).
__FILE__: The presumed name of the source file (a character string literal).
__DATE__: The date of translation of the source file (a character string literal...)
__TIME__: The time of translation of the source file (a character string literal...)
__STDC__: Whether__STDC__ is predefined
__cplusplus: The name __cplusplus is defined to the value 199711L when compiling a C ++ translation unit

So your code would be:

if(!Logical)
  printf("Not logical value at line number %d \n",__LINE__);

回答3:

You could use a macro with the same behavior as printf(), except that it also includes debug information such as function name, class, and line number:

#include <cstdio>  //needed for printf
#define print(a, args...) printf("%s(%s:%d) " a,  __func__,__FILE__, __LINE__, ##args)
#define println(a, args...) print(a "\n", ##args)

These macros should behave identically to printf(), while including java stacktrace-like information. Here's an example main:

void exampleMethod() {
    println("printf() syntax: string = %s, int = %d", "foobar", 42);
}

int main(int argc, char** argv) {
    print("Before exampleMethod()...\n");
    exampleMethod();
    println("Success!");
}

Which results in the following output:

main(main.cpp:11) Before exampleMethod()...
exampleMethod(main.cpp:7) printf() syntax: string = foobar, int = 42
main(main.cpp:13) Success!

回答4:

Use __LINE__ (that's double-underscore LINE double-underscore), the preprocessor will replace it with the line number on which it is encountered.

回答5:

Checkout __FILE__ and __LINE__ macros

回答6:

Try __FILE__ and __LINE__.
You might also find __DATE__ and __TIME__ useful.
Though unless you have to debug a program on the clientside and thus need to log these informations you should use normal debugging.

回答7:

Since i'm also facing this problem now and i cannot add an answer to a different but also valid question asked here, i'll provide an example solution for the problem of: getting only the line number of where the function has been called in C++ using templates.

Background: in C++ one can use non-type integer values as a template argument. This is different than the typical usage of data types as template arguments. So the idea is to use such integer values for a function call.

#include <iostream>

class Test{
    public:
        template<unsigned int L>
        int test(){
            std::cout << "the function has been called at line number: " << L << std::endl;
            return 0;
        }
        int test(){ return this->test<0>(); }
};

int main(int argc, char **argv){
    Test t;
    t.test();
    t.test<__LINE__>();
    return 0;
}

Output:

the function has been called at line number: 0

the function has been called at line number: 16

One thing to mention here is that in C++11 Standard it's possible to give default template values for functions using template. In pre C++11 default values for non-type arguments seem to only work for class template arguments. Thus, in C++11, there would be no need to have duplicate function definitions as above. In C++11 its also valid to have const char* template arguments but its not possible to use them with literals like __FILE__ or __func__ as mentioned here.

So in the end if you're using C++ or C++11 this might be a very interesting alternative than using macro's to get the calling line.

回答8:

C++20 offers a new way to achieve this by using std::source_location. This is currently accessible in as std::experimental::source_location.

The problem with macros like __LINE__ is that if you want to create a, for example, a logging function that outputs the current line number along with a message, you always have to pass __LINE__ as a function argument, because it is expanded at the call site. Something like this:

void log(const std::string msg) {
    std::cout << __LINE__ << " " << msg << std::endl;
}

Will always output the line of the function declaration and not the line where log was actually called. On the other hand, with std::source_location you can write something like this:

void log(const std::string msg,
  const std::source_location loc = std::experimental::source_location::current()) {
    std::cout << loc.line() << " " << msg << std::endl;
}

Here, loc is initialized with the line number pointing to the location where log was called.

来源：https://stackoverflow.com/questions/2849832/c-c-line-number

标签

c++

c-preprocessor

line-numbers

dynamic-compilation