I'm writing c++ program using codeblocks IDE
int main()
{
int i =0;
int f = 3.14;
i = f; //must give me a warning message, possible loss data.
}
Why the compilation not show a narrowing warning message?
How to enable that?
Note: I have fixed my compiler options as -std=c++11 -Wall
in the other compiler options put -Wconversion
( code::block 16 )
for:
int i =0;
int f = 3.14;
i = f;
warning: conversion to ‘int’ alters ‘double’ constant
value [-Wfloat-conversion]
Some useful warning that I use always:
-Wall -Weffc++ -Wextra -pedantic -Wfatal-errors -pedantic-errors
Related
#include <iostream>
using namespace std;
int main() {
unsigned int u = 5;
int x = -1;
if(x>u) {
cout<<"Should not happen"<<endl;
} else {
cout<<"Ok"<<endl;
}
}
This code outputs Should not happen. I came across this when comparing the size of a string (size_t is an unsigned int or unsigned long long value) to an int.
It seems that C type casts int to unsigned int but in practice, it seems it would bring in bugs. Honestly, I would have preferred a compile-time error given how incompatible int is to unsigned int. I would like to know why the convention is like this?
You can enable -Wsign-compare -Werror in Clang: Try it online!
It'll produce a compile-time error (because of -Werror that treats warnings as errors):
.code.tio.cpp:7:9: error: comparison of integers of different signs: 'int' and 'unsigned int' [-Werror,-Wsign-compare]
if(x>u) {
~^~
1 error generated.
For some reason, -Wall -Werror in Clang (but not in GCC) doesn't produce any errors. But -Wall -Wextra -Werror does include -Wsign-compare, so you can use that.
In the following snippet no warnings are produced. g++4.4.3 -Wall -pedantic
//f is
void f(int );
f(3.14);
double d = 3.14;
int i = d+2;
I have a strong recollection of this being a warning, something along the lines of "Possible loss of precision". Was it removed or is my memory playing tricks on me?
How can i turn this into a warning in g++? I find this a useful warning, or is it a bad idea?
I can't even find anything appropriate at http://gcc.gnu.org/onlinedocs/gcc-4.4.5/gcc/Warning-Options.html
$ gcc -Wconversion test.c
test.c: In function ‘main’:
test.c:3: warning: conversion to ‘int’ from ‘double’ may alter its value
Use -Wconversion option. -Wall doesn't include it.
With -Wconversion option, GCC gives these warning messages:
warning: conversion to 'int' alters 'double' constant value
warning: conversion to 'int' from 'double' may alter its value
Apart from what other answers mention it is also worth mentioning that in C++0x {} initialization doesn't narrow. So instead of getting a warning you'll get an error for example
void f(int x)
{
// code
}
int main()
{
f({3.14}); // narrowing conversion of '3.14000000000000012434497875801753252744674682617e+0' from 'double' to 'int' inside { }
}
g++ 4.4 and above support initializer list (with -std=c++0x option)
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.
In the following snippet no warnings are produced. g++4.4.3 -Wall -pedantic
//f is
void f(int );
f(3.14);
double d = 3.14;
int i = d+2;
I have a strong recollection of this being a warning, something along the lines of "Possible loss of precision". Was it removed or is my memory playing tricks on me?
How can i turn this into a warning in g++? I find this a useful warning, or is it a bad idea?
I can't even find anything appropriate at http://gcc.gnu.org/onlinedocs/gcc-4.4.5/gcc/Warning-Options.html
$ gcc -Wconversion test.c
test.c: In function ‘main’:
test.c:3: warning: conversion to ‘int’ from ‘double’ may alter its value
Use -Wconversion option. -Wall doesn't include it.
With -Wconversion option, GCC gives these warning messages:
warning: conversion to 'int' alters 'double' constant value
warning: conversion to 'int' from 'double' may alter its value
Apart from what other answers mention it is also worth mentioning that in C++0x {} initialization doesn't narrow. So instead of getting a warning you'll get an error for example
void f(int x)
{
// code
}
int main()
{
f({3.14}); // narrowing conversion of '3.14000000000000012434497875801753252744674682617e+0' from 'double' to 'int' inside { }
}
g++ 4.4 and above support initializer list (with -std=c++0x option)
In the following snippet no warnings are produced. g++4.4.3 -Wall -pedantic
//f is
void f(int );
f(3.14);
double d = 3.14;
int i = d+2;
I have a strong recollection of this being a warning, something along the lines of "Possible loss of precision". Was it removed or is my memory playing tricks on me?
How can i turn this into a warning in g++? I find this a useful warning, or is it a bad idea?
I can't even find anything appropriate at http://gcc.gnu.org/onlinedocs/gcc-4.4.5/gcc/Warning-Options.html
$ gcc -Wconversion test.c
test.c: In function ‘main’:
test.c:3: warning: conversion to ‘int’ from ‘double’ may alter its value
Use -Wconversion option. -Wall doesn't include it.
With -Wconversion option, GCC gives these warning messages:
warning: conversion to 'int' alters 'double' constant value
warning: conversion to 'int' from 'double' may alter its value
Apart from what other answers mention it is also worth mentioning that in C++0x {} initialization doesn't narrow. So instead of getting a warning you'll get an error for example
void f(int x)
{
// code
}
int main()
{
f({3.14}); // narrowing conversion of '3.14000000000000012434497875801753252744674682617e+0' from 'double' to 'int' inside { }
}
g++ 4.4 and above support initializer list (with -std=c++0x option)