Cannot implicitly convert type int to short

Question

I wrote the following small program to print out the Fibonacci sequence   static void Main string   args        Console Write  Please give a value for n         Int16 n   Int16 Parse Console ReadLine          Int16 firstNo   0      Int16 secondNo   1       Console WriteLine firstNo       Console WriteLine secondNo        for  Int16 i   0  i  lt  n  i                    Problem on this line                             Int16 answer   firstNo   secondNo           Console WriteLine answer            firstNo   secondNo          secondNo   answer             Console ReadLine         The compilation message is      Cannot implicitly convert type  int    to  short   An explicit conversion   exists  are you missing a cast     Since everything involved is an Int16  short  then why are there any implicit conversions going on  And more specificially why the failure here  and not when initially assigning an int to the variable    An explanation would be much appreciated

User · Answer

For some strange reason  you can use the    operator to add shorts   short answer   0  short firstNo   1  short secondNo   2   answer    firstNo  answer    secondNo

User · Answer

The line   Int16 answer   firstNo   secondNo    is interpreted as   Int16 answer    Int32   firstNo   secondNo     Simply because there is no such thing as Int16 arithmetic    The simple solution  Do not use Int16  Use Int32 or simply int   int is your default integer type  short and long are only used in special cases

User · Answer

The problem is  that adding two Int16 results in an Int32 as others have already pointed out  Your second question  why this problem doesn t already occur at the declaration of those two variables is explained here  http   msdn microsoft com en-us library ybs77ex4 28v VS 71 29 aspx   short x   32767   In the preceding declaration  the integer literal 32767 is implicitly converted from int to short  If the integer literal does not fit into a short storage location  a compilation error will occur   So  the reason why it works in your declaration is simply that the literals provided are known to fit into a short

User · Answer

The result of summing two Int16 variables is an Int32   Int16 i1   1  Int16 i2   2  var result   i1   i2  Console WriteLine result GetType   Name     It outputs Int32

User · Answer

Adding two Int16 values result in an Int32 value  You will have to cast it to Int16   Int16 answer    Int16   firstNo   secondNo     You can avoid this problem by switching all your numbers to Int32

User · Answer

Read Eric Lippert  s answers to these questions   Integer summing blues  short    short problem Unary minus on a short becomes an int

User · Answer

The plus operator converts operands to int first and then does the addition  So the result is the int  You need to cast it back to short explicitly because conversions from a  longer  type to  shorter  type a made explicit  so that you don t loose data accidentally with an implicit cast   As to why int16 is cast to int  the answer is  because this is what is defined in C  spec  And C  is this way is because it was designed to closely match to the way how CLR works  and CLR has only 32 64 bit arithmetic and not 16 bit  Other languages on top of CLR may choose to expose this differently

User · Answer

That s because the result of adding two Int16 is an Int32  Check the  conversions  paragraph here  http   msdn microsoft com en-us library ybs77ex4 28v vs 71 29 aspx

User · Answer

Microsoft converts your Int16 variables into Int32 when doing the add function   Change the following   Int16 answer   firstNo   secondNo    into     Int16 answer    Int16  firstNo   secondNo

[c#] Cannot implicitly convert type 'int' to 'short'

Examples related to c#

Examples related to int

Examples related to short