After a long time trace of my program I finally found that abs is the blameable part of my program. What should I expect from this code? why do I get:
x=0.1
|x|=0
#include <iostream>
int main()
{
double x=0.1;
std::cout<<"x="<<x<<std::endl;
std::cout<<"|x|="<<abs(x)<<std::endl;
return 0;
}
You may be wondering "But why I didnt get warning on g++ -g -Wall -Wfatal-errors -Wextra -std=c++11 test.cpp -o ./bin/test -lboost_filesystem -lboost_system?"
Turns out Wall isn't quite "all".
g++ -g -Wconversion -std=c++11 test.cpp -o tester -lboost_filesystem -lboost_system
test.cpp: In function ‘int main()’:
test.cpp:7:29: warning: conversion to ‘int’ from ‘double’ may alter its value [-Wconversion]
std::cout<<"|x|="<<abs(x)<<std::endl;
^
clang-3.6's diagnostic is clearer still, and no explicit opt-in required:
$ clang++ -std=c++11 test.cpp -o tester
test.cpp:8:24: warning: using integer absolute value function 'abs' when argument is of floating point type [-Wabsolute-value]
std::cout<<"|x|="<<abs(x)<<std::endl;
^
test.cpp:8:24: note: use function 'std::abs' instead
std::cout<<"|x|="<<abs(x)<<std::endl;
^~~
std::abs
test.cpp:8:24: note: include the header <cmath> or explicitly provide a declaration for 'std::abs'
1 warning generated.
You're using the abs defined in <cstdlib>, which only works on integers.
Use the abs defined in <cmath> instead. It works on floating point values.
Related
My environment Arch Linux, gcc 7.2
I'm learning C++ and I'm using keyword constexpr to define a constant, while compile, it give me an error message
error: identifier ‘constexpr’ is a keyword in C++11 [-Werror=c++11-compat]
I can compile my program with default g++, but cannot compile with -std=c++14 and -Werror
The command I'm using is:
g++ -std=c++14 -O2 -Wall -Werror -Wextra -ansi -flto
I believe the -Werror option caused the issue. but what is the issue? can someone tell me please?
#include <iostream>
int main() {
constexpr double yen_dollar = 0.107;
std::cout << yen_dollar << std::endl;
return 0;
}
test.cpp:4:5: error: identifier ‘constexpr’ is a keyword in C++11 [-Werror=c++11-compat]
constexpr double yen_dollar = 0.107;
^~~~~~~~~
test.cpp: In function ‘int main()’:
test.cpp:4:5: error: ‘constexpr’ was not declared in this scope
test.cpp:5:16: error: ‘yen_dollar’ was not declared in this scope
std::cout << yen_dollar << std::endl;
From the GCC documentation §3.4 Options Controlling C Dialect, one can read:
-ansi
In C mode, this is equivalent to -std=c90. In C++ mode, it is equivalent to -std=c++98.
Since, you compiled with
g++ -std=c++14 -O2 -Wall -Werror -Wextra -ansi -flto
-ansi overwrites -std=c++14 with -std=c++98. This is why constexpr is not recognized.
Solution: get rid of the -ansi flag.
GCC will helpfully warn you if you forget to include the NULL sentinel at the end of a call to one of the exec(3) functions:
#include <unistd.h>
int main(int argc, char **argv)
{
execlp("test", "test", "arg1");
}
Sample compiler output for GCC 4.8:
$ g++ test.cc -Wformat
test.cc: In function ‘int main(int, char**)’:
test.cc:4:32: warning: missing sentinel in function call [-Wformat=]
execlp("test", "test", "arg1");
^
$
However, if you compile in C++11 mode, no diagnostic is printed:
$ g++ test.cc -std=c++11 -Wformat
$
Why is this warning not available in C++11? Is there any way to get it back?
execlp is not a standard C function. For the compiler to recognize it as a "standard" function, for which it knows what the arguments should look like, you need -std=gnu++11 instead of -std=c++11. Note that the default is -std=gnu++98. Glibc could improve the situation by specifying the sentinel attribute on the declaration of execlp.
int a;
cin >> a;
int ints[a];
Why does this not throw any kind of warning while compiling? How do I know when this array thing is actually using the heap or the stack?
g++ -std=c++11 -Wall *.cpp -o main
ISO C++ disallows the use of variable length arrays, which g++ happily tells you if you increase the strictness of it by passing it the -pedantic flag.
Using -pedantic will issue a warning about things breaking the standard. If you want g++ to issue an error and with this refuse compilation because of such things; use -pedantic-errors.
g++ -Wall -pedantic -std=c++11 apa.cpp
apa.cpp: In function ‘int main(int, char**)’:
apa.cpp:8:13: warning: ISO C++ forbids variable length array ‘ints’ [-Wvla]
int ints[a];
^
apa.cpp:8:7: warning: unused variable ‘ints’ [-Wunused-variable]
int ints[a];
^
I've installed GCC 4.8 using this method on my Mac. Everything works fine except that for certain functions like scanf and printf, the program compiles fine without any error/warning even when I did not include their respective libraries like cstdio. Is there any way that I can do to for GCC (more specifically G++, as I am dealing with C++ programs) to throw an error when such code is being fed? The following code compiles fine on my machine:
#include <iostream>
//Notice I did not include cstdio but my program uses printf later on
int main()
{
printf("Hello World!\n");
return 0;
}
I was given the suggestion to use -Werror-implicit-function-declaration -Werror or -Wall -Werror, but they don't work.
-Wimplicit-function-declaration -Werror works for me. There must be some other problems as well.
h2co3-macbook:~ h2co3$ cat baz.c
#ifndef BAILZ_OUT
#include <stdio.h>
#endif
int main()
{
printf("Hello world!\n");
return 0;
}
h2co3-macbook:~ h2co3$ gcc -o baz baz.c -Wimplicit-function-declaration -Werror
h2co3-macbook:~ h2co3$ echo $?
0
h2co3-macbook:~ h2co3$ gcc -o baz baz.c -Wimplicit-function-declaration -Werror -DBAILZ_OUT
cc1: warnings being treated as errors
baz.c: In function ‘main’:
baz.c:7: warning: implicit declaration of function ‘printf’
baz.c:7: warning: incompatible implicit declaration of built-in function ‘printf’
h2co3-macbook:~ h2co3$ echo $?
1
h2co3-macbook:~ h2co3$
The reason you get no diagnostic is that <iostream> is including the declaration of printf, which it seems to do with the c++0x or c++11 flags.
This compiles on a gcc 4.8 snapshot with the following command line:
g++ -Wall -Wextra -pedantic-errors -std=c++0x
#include <iostream>
int main()
{
printf("Hello World!\n");
return 0;
}
If you comment out the <iostream> include, or remove the C++11 compilation flags, you get an error.
impl_decl.cpp: In function 'int main()':
impl_decl.cpp:5:28: error: 'printf' was not declared in this scope
From the Annex C/Compatibility of the C++ standard from 2003:
C.1 C++ and ISO C:
C.1.3 Clause 5: expressions [diff.expr]
5.2.2
Change: Implicit declaration of functions is not allowed
Rationale: The type-safe nature of C++.
That means that implicit declarations must cause a compilation error in C++.
I'm guessing you're compiling not C++ files, but C files and you're doing that in some pre-C99 mode, which is the default in gcc. The C standard from 1999 disallows implicit declarations as well.
You may want to pass to gcc a combination of these options: -std=c99 -Werror.
Suppose we have the following code:
#if !defined(__cplusplus)
# error This file should be compiled as C++
#endif
#include <stdio.h>
#include <string>
//#define USE_CXX_CLASS
#ifdef USE_CXX_CLASS
class SomeClass
{
public:
SomeClass() {}
~SomeClass() {}
std::string GetSomeString()
{
// case #1
}
};
#endif // USE_CXX_CLASS
int foo()
{
// case #2
}
int
main (int argc, char *argv[])
{
(void)argc;
(void)argv;
#ifdef USE_CXX_CLASS
SomeClass someInstance;
someInstance.GetSomeString();
#endif // USE_CXX_CLASS
foo();
return 0;
}
And suppose that it were to be compiled the C++ compiler (and not the C compiler) from GCC version 4.2.1 with the options -Wreturn-type -Werror=return-type. If the above code is compiled as is without first uncommenting the //#define USE_CXX_CLASS line above, then you will see a warning but no error:
.../gcc-4.2.1/bin/g++ -g -fPIC -Wreturn-type -Werror=return-type test.cpp -c -o test.o
test.cpp: In function 'int foo()':
test.cpp:26: warning: control reaches end of non-void function
But if the //#define USE_CXX_CLASS line is uncommented, then the warning is treated as an error:
.../gcc-4.2.1/bin/g++ -g -fPIC -Wreturn-type -Werror=return-type test.cpp -c -o test.o
test.cpp: In member function 'std::string SomeClass::GetSomeString()':
test.cpp:18: error: no return statement in function returning non-void [-Wreturn-type]
gmake: *** [test.o] Error 1
Yes, one is a non-member function (case #2), and the other is a C++ function (case #1). IMO, that should not matter. I want both conditions treated as an error, and I don't want to add -Werror or -Wall at this point in time (probably will do so later, but that is out of scope of this question).
My sub-questions are:
Is there some GCC switch that I am missing that should work? (No I do not want to use #pragma's.)
Is this a bug that has been addressed in a more recent version of GCC?
For reference, I have already poured through other similar questions already, including the following:
Why does flowing off the end of a non-void function without returning a value not produce a compiler error?
C question: no warning?
Is a return statement mandatory for C++ functions that do not return void?
It has been fixed, it works well with g++ 9.3: both member functions and free functions are treated as error with -Wall -Werror=return-type
I do see an error even w/o the USE_CXX_CLASS flag. i.e. g++ is consistent with the error for both class member functions and non member functions.
g++ (GCC) 4.4.3 20100127 (Red Hat 4.4.3-4)
It seems to me that what you need is a shell script wrapper around gcc.
Name it something like gcc-wrapper and g++-wrapper.
In your Makefile set CC and CXX to the wrappers.
Have the wrapper invoke GCC and pipe its output to another program which will search for your desired warning strings.
Have the search program exit with an error when it finds the warning.