Complex literal 'i' used in function argument

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There seems to be a problem, using the literal i in C++ with std::complex.



Consider the following code:



std::complex<double> a = -1.0i * 42.0;
std::complex<double> b = a + 1.0i;


The second line fails to compile with:
error: no match for ‘operator+’ (operand types are ‘std::complex<double>’ and ‘__complex__ double’)



This also shows up when using the complex literal in function calls, e.g.



std::exp<std::complex<double>>( 1.0i * 3.14159 );


How come the complex literal 1.0i is not convertible to std::complex<double>?



Do I have to explicitly construct a std::complex with 1.0i?




c++ std complex-numbers




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  • 3




    Which compiler and version? This compiles for me on GCC trunk, Clang trunk, and a slightly older version of MSVC.
    – chris
    Aug 8 at 9:06







  • 1




    For example, ideone's gcc 6.3, as shown here : ideone.com/qxw5eI
    – Louen
    Aug 8 at 9:09






  • 1




    Somehow it compiles fine on wandbox's gcc 6.3
    – VTT
    Aug 8 at 9:11















up vote
20
down vote

favorite












There seems to be a problem, using the literal i in C++ with std::complex.



Consider the following code:



std::complex<double> a = -1.0i * 42.0;
std::complex<double> b = a + 1.0i;


The second line fails to compile with:
error: no match for ‘operator+’ (operand types are ‘std::complex<double>’ and ‘__complex__ double’)



This also shows up when using the complex literal in function calls, e.g.



std::exp<std::complex<double>>( 1.0i * 3.14159 );


How come the complex literal 1.0i is not convertible to std::complex<double>?



Do I have to explicitly construct a std::complex with 1.0i?




c++ std complex-numbers




share|improve this question


















  • 3




    Which compiler and version? This compiles for me on GCC trunk, Clang trunk, and a slightly older version of MSVC.
    – chris
    Aug 8 at 9:06







  • 1




    For example, ideone's gcc 6.3, as shown here : ideone.com/qxw5eI
    – Louen
    Aug 8 at 9:09






  • 1




    Somehow it compiles fine on wandbox's gcc 6.3
    – VTT
    Aug 8 at 9:11













up vote
20
down vote

favorite









up vote
20
down vote

favorite











There seems to be a problem, using the literal i in C++ with std::complex.



Consider the following code:



std::complex<double> a = -1.0i * 42.0;
std::complex<double> b = a + 1.0i;


The second line fails to compile with:
error: no match for ‘operator+’ (operand types are ‘std::complex<double>’ and ‘__complex__ double’)



This also shows up when using the complex literal in function calls, e.g.



std::exp<std::complex<double>>( 1.0i * 3.14159 );


How come the complex literal 1.0i is not convertible to std::complex<double>?



Do I have to explicitly construct a std::complex with 1.0i?




c++ std complex-numbers




share|improve this question














There seems to be a problem, using the literal i in C++ with std::complex.



Consider the following code:



std::complex<double> a = -1.0i * 42.0;
std::complex<double> b = a + 1.0i;


The second line fails to compile with:
error: no match for ‘operator+’ (operand types are ‘std::complex<double>’ and ‘__complex__ double’)



This also shows up when using the complex literal in function calls, e.g.



std::exp<std::complex<double>>( 1.0i * 3.14159 );


How come the complex literal 1.0i is not convertible to std::complex<double>?



Do I have to explicitly construct a std::complex with 1.0i?





c++ std complex-numbers





share|improve this question













share|improve this question




share|improve this question








edited Aug 8 at 14:24









Peter Mortensen

12.9k1983111




12.9k1983111










asked Aug 8 at 9:03









Louen

1,99311430




1,99311430







  • 3




    Which compiler and version? This compiles for me on GCC trunk, Clang trunk, and a slightly older version of MSVC.
    – chris
    Aug 8 at 9:06







  • 1




    For example, ideone's gcc 6.3, as shown here : ideone.com/qxw5eI
    – Louen
    Aug 8 at 9:09






  • 1




    Somehow it compiles fine on wandbox's gcc 6.3
    – VTT
    Aug 8 at 9:11













  • 3




    Which compiler and version? This compiles for me on GCC trunk, Clang trunk, and a slightly older version of MSVC.
    – chris
    Aug 8 at 9:06







  • 1




    For example, ideone's gcc 6.3, as shown here : ideone.com/qxw5eI
    – Louen
    Aug 8 at 9:09






  • 1




    Somehow it compiles fine on wandbox's gcc 6.3
    – VTT
    Aug 8 at 9:11








3




3




Which compiler and version? This compiles for me on GCC trunk, Clang trunk, and a slightly older version of MSVC.
– chris
Aug 8 at 9:06





Which compiler and version? This compiles for me on GCC trunk, Clang trunk, and a slightly older version of MSVC.
– chris
Aug 8 at 9:06





1




1




For example, ideone's gcc 6.3, as shown here : ideone.com/qxw5eI
– Louen
Aug 8 at 9:09




For example, ideone's gcc 6.3, as shown here : ideone.com/qxw5eI
– Louen
Aug 8 at 9:09




1




1




Somehow it compiles fine on wandbox's gcc 6.3
– VTT
Aug 8 at 9:11





Somehow it compiles fine on wandbox's gcc 6.3
– VTT
Aug 8 at 9:11













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You should recompile with --std=c++14 (no GNU ext) to avoid conflict of i suffix with gcc extension




The ISO C++14 library also defines the ‘i’ suffix, so C++14 code that includes the <complex> header cannot use ‘i’ for the GNU extension. The ‘j’ suffix still has the GNU meaning.







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    active

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    up vote
    24
    down vote



    accepted










    You should recompile with --std=c++14 (no GNU ext) to avoid conflict of i suffix with gcc extension




    The ISO C++14 library also defines the ‘i’ suffix, so C++14 code that includes the <complex> header cannot use ‘i’ for the GNU extension. The ‘j’ suffix still has the GNU meaning.







    share|improve this answer


























      up vote
      24
      down vote



      accepted










      You should recompile with --std=c++14 (no GNU ext) to avoid conflict of i suffix with gcc extension




      The ISO C++14 library also defines the ‘i’ suffix, so C++14 code that includes the <complex> header cannot use ‘i’ for the GNU extension. The ‘j’ suffix still has the GNU meaning.







      share|improve this answer
























        up vote
        24
        down vote



        accepted







        up vote
        24
        down vote



        accepted






        You should recompile with --std=c++14 (no GNU ext) to avoid conflict of i suffix with gcc extension




        The ISO C++14 library also defines the ‘i’ suffix, so C++14 code that includes the <complex> header cannot use ‘i’ for the GNU extension. The ‘j’ suffix still has the GNU meaning.







        share|improve this answer














        You should recompile with --std=c++14 (no GNU ext) to avoid conflict of i suffix with gcc extension




        The ISO C++14 library also defines the ‘i’ suffix, so C++14 code that includes the <complex> header cannot use ‘i’ for the GNU extension. The ‘j’ suffix still has the GNU meaning.








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Aug 8 at 9:18

























        answered Aug 8 at 9:12









        VTT

        21.1k32143




        21.1k32143



























             

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