C language The problem with the use of semicolon in C language. Thank you very much!

The semicolon is the sign of the end of the statement, but the semicolon is not used after the loop, if, and subfunctions such as long long a (int b), and everything else is used, but two semicolons cannot be added, although it will not cause an error, but it may affect the result.

It's a sign of the end of a statement.,Some sentences are not added because the statement hasn't ended!

At the end of all statements, add ;

Prototype: int winapi icepub dictionarycodetransfer2(char *strdictionaryfilename, char *strsrc, char *strcode, char *strfenge).

Enter: strdictionaryfilename dictionary filename.

strsrc words to be processed.

The separator string between a word and a code in a strfenge dictionary.

Output: strcode strsrc information.

return code: char strcode[1024];

typedef int (winapi icepub_dictionarycodetransfer2)(char *strdictionaryfilename,char *strsrc,char *strcode,char *strfenge);

icepub_dictionarycodetransfer2 *icepub_dictionarycodetransfer2 = 0;

hinstance hdlldrv = loadlibrary("");

if(hdlldrv)

icepub_dictionarycodetransfer2 = (icepub_dictionarycodetransfer2 *)getprocaddress(hdlldrv, "icepub_dictionarycodetransfer2");

if(icepub_dictionarycodetransfer2)

icepub_dictionarycodetransfer2("Chinese-English Dictionary. txt","Me",strcode, " ");

if(hdlldrv)

freelibrary(hdlldrv);

afxmessagbox(strcode);

In fact, the semicolon has the same function as the tonal sign, except that the semicolon is used to separate sentences, and the semicolon is used to separate words!

A semicolon is used when a program statement is temporarily terminated.

All statements in the function must be added, except for those with {}.

Seems like that.

It's the same problem, correct way to write it: e must be preceded by a number, and e must not be followed by a decimal.

It is specified that characters and numbers are common in the ASCII range, but null is predefined as 0 and is not empty.

4.Resend it, I can't see it clearly.

5.'\137', c provisions' 'Within is a character, followed by an octal number, and followed by a hexadecimal number.

Because the comma operator has the lowest priority (nothing lower), the final result is a=1 b=0 c=1

Calculate a=b=c=0 first, so that abc is 0

Then calculate a++&b||c++

Since a++ is a value that is used first and then ++ is calculated, it is false (i.e. 0), which causes ++b to not be executed.

Finally, C++ is executed

So the result is as shown above.