Error invalid operands of types const char 35 and const char 2 to binary operator

Question

At the top of my file I have   define AGE  42    Later in the file I use ID multiple times including some lines that look like  1 std  string name    Obama   2 std  string str    Hello     name     you are     AGE     years old    3 str     Do you feel     AGE     years old      I get the error       error  invalid operands of types    const char  35     and    const char  2     to binary    operator        on line 3  I did some research and found it was because of how C   was treating the different strings and was able to fix it by changing  AGE  to  string AGE    However  I accidentally missed one of the instances until today and was wondering why the compiler wasn t complaining even though I still had an instance where it was just  AGE    Through some trial and error I found that I only need string AGE  on lines where I don t concatenate another string that was created in the function body   My questions is  what is going on in the background that C   doesn t like concatenating a string with a string put there by the preprocessor unless you are also concatenating string that you defined in the function

User · Answer

In this particular case  an even simpler fix would be to just get rid of the     all together because AGE is a string literal and what comes before and after are also string literals   You could write line 3 as   str     Do you feel   AGE   years old     This is because most C C   compilers will concatenate string literals automatically   The above is equivalent to   str     Do you feel    42    years old     which the compiler will convert to   str     Do you feel 42 years old

User · Answer

Consider this   std  string str    Hello      world      bad    Both the rhs and the lhs for operator   are char s  There is no definition of operator   that takes two char s  in fact  the language doesn t permit you to write one   As a result  on my compiler this produces a  cannot add two pointers  error  yours apparently phrases things in terms of arrays  but it s the same problem    Now consider this   std  string str    Hello     std  string  world       ok   There is a definition of operator   that takes a const char  as the lhs and a std  string as the rhs  so now everyone is happy   You can extend this to as long a concatenation chain as you like  It can get messy  though  For example   std  string str    Hello      there     std  string  world       no good    This doesn t work because you are trying to   two char s before the lhs has been converted to std  string  But this is fine   std  string str   std  string  Hello       there      world      ok   Because once you ve converted to std  string  you can   as many additional char s as you want   If that s still confusing  it may help to add some brackets to highlight the associativity rules and then replace the variable names with their types     std  string  Hello       there       world      string   char     char     The first step is to call string operator  string  char    which is defined in the standard library  Replacing those two operands with their result gives     string    char     Which is exactly what we just did  and which is still legal  But try the same thing with     char    char     string    And you re stuck  because the first operation tries to add two char s   Moral of the story  If you want to be sure a concatenation chain will work  just make sure one of the first two arguments is explicitly of type std  string

User · Answer

You cannot concatenate raw strings like this  operator  only works with two std  string objects or with one std  string and one raw string  on either side of the operation    std  string s         s   s     OK s    x      OK  x    s     OK  x     x     error   The easiest solution is to turn your raw string into a std  string first    Do you feel     std  string AGE      years old      Of course  you should not use a macro in the first place  C   is not C  Use const or  in C  11 with proper compiler support  constexpr

User · Answer

AGE is defined as  42  so the line   str     Do you feel     AGE     years old      is converted to   str     Do you feel      42      years old      Which isn t valid since  Do you feel   and  42  are both const char    To solve this  you can make one a std  string  or just remove the        1  str    std  string  Do you feel      AGE     years old        2  str     Do you feel   AGE   years old

User · Answer

In line 2  there s a std  string involved  name   There are operations defined for char     std  string  std  string   char    etc   Hello     name gives a std  string  which is added to   you are    giving another string  etc   In line 3  you re saying  char     char     char     and you can t just add arrays to each other

User · Answer

I had the same problem in my code  I was concatenating a string to create a string  Below is the part of code   int scannerId   1  std strring testValue  strInXml   std  string std  string   lt inArgs gt                               lt scannerID gt     scannerId    std  string   lt  scannerID gt                               lt cmdArgs gt                               lt arg-string gt     testValue      lt  arg-string gt                               lt arg-bool gt FALSE lt  arg-bool gt                               lt arg-bool gt FALSE lt  arg-bool gt                               lt  cmdArgs gt                              lt  inArgs gt

[c++] Error: invalid operands of types ‘const char [35]’ and ‘const char [2]’ to binary ‘operator+’

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