一种情况

5）不，开玩笑吧！说真的这次： char GetSortMethod（int sortMeth） { if（sortMeth == 0） 返回''D''; if（sortMeth == 1） 返回''A''; 返回''\'''; } 6）原来的功能是这样使用的： printf（somevar =''％s''\ n，GetSortMethod（anum））; / * int anum; * / 这是使用正确版本的最佳方式吗？ char meth [2]; meth [0] = GetSortMethod（anum）; meth [1] =''\ 0''; printf（" somevar = ''％s''\ n，meth）; < OT> 7）原始函数实现在？ 8）是否存在原始函数的可想象的情况 实现可能是正确的？即使它真的是一个C ++ 类成员函数？ < / OT> 9）任何其他建议还是挑剔？ - Christopher Benson-Manica |我*应该*知道我在说什么 - 如果我 ataru（at）cyberspace.org |不，我需要知道。火焰欢迎。(I did not write this code!)const char *GetSortMethod( unsigned int sortMeth ){const char *sortingstring;if( sortMeth == 0 ) {sortingstring = "D";}if( sortMeth == 1 ) {sortingstring = "A";}if( sortMeth == -1 ) {sortingstring = "";}return( sortingstring );}1) Returning an automatic pointer to a string literal is a bigmistake, correct?2) If sortMeth isn''t -1, 0, or 1, the value returned ieven less determinate than it was before...3) sortMeth is *never* -1, since it''s unsigned...4) How about the following correction?char GetSortMethod( int sortMeth ){return( !sortMeth ? ''D'' : sortMeth == 1 ? ''A'' : 0 );}5) No, just kidding! Seriously this time:char GetSortMethod( int sortMeth ){if( sortMeth == 0 )return ''D'';if( sortMeth == 1 )return ''A'';return ''\0'';}6) The original function was used like this:printf( "somevar=''%s''\n", GetSortMethod(anum) ); /* int anum; */Is this the best way to use the correct version?char meth[2];meth[0]=GetSortMethod( anum );meth[1]=''\0'';printf( "somevar=''%s''\n", meth );<OT>7) Is the original function implementation any more viable whencompiled by a C++ compiler (namely, Borland C++ Builder 4.0)?8) Is there any conceivable circumstance where the original functionimplementation could have been correct? Even if it''s really a C++class member function?</OT>9) Any other suggestions or nitpicks?--Christopher Benson-Manica | I *should* know what I''m talking about - if Iataru(at)cyberspace.org | don''t, I need to know. Flames welcome.推荐答案 Christopher Benson-Manica< at *** @ nospam.cyberspace.org>潦草地写道：Christopher Benson-Manica <at***@nospam.cyberspace.org> scribbled the following:（我没有写这段代码！） const char * GetSortMethod（unsigned int sortMeth） {char /> const char * sortingstring; if（sortMeth == 0）{ sortingstring =" D" ;; } if（sortMeth == 1）{ sortingstring =" A"; } if（sortMeth == -1）{ sortingstring ="" ;; } return（sortingstring）; } 1）返回一个字符串文字的自动指针是一个很大的错误，对吗？ 编号字符串文字的存储空间可以在执行 整个程序后继续存在。以上代码的安全性不低于： int GetInt（void）{ int i; i = 1; 返回i; } 2）如果sortMeth不是-1，0或1，则返回的值更加确定比以前... 嗯，这是不确定的。在返回函数之前说返回值 是没有意义的。 3）sortMeth是* never * -1，因为它是未签名的。 .. 正确。 4）以下更正怎么样？ char GetSortMethod（int sortMeth） {返回（！sortMeth？''D''：sortMeth == 1？''A''：0）; } 5）不，开个玩意儿！严肃地说这次： char GetSortMethod（int sortMeth） { if（sortMeth == 0）返回''D''; if（sortMeth == 1）返回''A''; 返回''\ 0''; } 这样可行。 6）原始函数的使用方式如下： printf（" somevar =''％s''\ n，GetSortMethod（anum））; / * int anum; * / 这是使用正确版本的最佳方式吗？ char meth [2]; meth [0] = GetSortMethod（anum）; meth [1] =''\'''; printf （somevar =''％s''\ n，meth）; 为什么不简单呢？ printf（" somevar =''％c''\ n，GetSortMethod（anum） ）; 9）任何其他建议或挑剔？ (I did not write this code!) const char *GetSortMethod( unsigned int sortMeth ) { const char *sortingstring; if( sortMeth == 0 ) { sortingstring = "D"; } if( sortMeth == 1 ) { sortingstring = "A"; } if( sortMeth == -1 ) { sortingstring = ""; } return( sortingstring ); } 1) Returning an automatic pointer to a string literal is a big mistake, correct?No. String literals have storage that survives the execution of thewhole program. The above code is no less safe than:int GetInt(void) {int i;i=1;return i;} 2) If sortMeth isn''t -1, 0, or 1, the value returned i even less determinate than it was before...Well, it''s indeterminate. It makes no sense to speak of the return valuebefore the return of the function. 3) sortMeth is *never* -1, since it''s unsigned...Right. 4) How about the following correction? char GetSortMethod( int sortMeth ) { return( !sortMeth ? ''D'' : sortMeth == 1 ? ''A'' : 0 ); } 5) No, just kidding! Seriously this time: char GetSortMethod( int sortMeth ) { if( sortMeth == 0 ) return ''D''; if( sortMeth == 1 ) return ''A''; return ''\0''; }That would work. 6) The original function was used like this: printf( "somevar=''%s''\n", GetSortMethod(anum) ); /* int anum; */ Is this the best way to use the correct version? char meth[2]; meth[0]=GetSortMethod( anum ); meth[1]=''\0''; printf( "somevar=''%s''\n", meth );Why not simply this?printf("somevar=''%c''\n", GetSortMethod(anum)); 9) Any other suggestions or nitpicks? 我无能为力。 - / - Joona Palaste（pa*****@cc.helsinki.fi）-------------芬兰----- --- \ \ - http：// www.helsinki.fi/~palaste ---------------------规则！ -------- / 正常就是其他人都是，而你却不是。 - Tolian Soran博士None I can think of.--/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/"Normal is what everyone else is, and you''re not."- Dr. Tolian Soran Joona I Palaste< pa ***** @ cc.helsinki.fi>这样说了：Joona I Palaste <pa*****@cc.helsinki.fi> spoke thus:否。字符串文字的存储在整个程序执行后仍然存在。 知道了。 为什么不简单呢？ printf（" somevar =''％c''\ n"，GetSortMethod（anum））; No. String literals have storage that survives the execution of the whole program.Got it. Why not simply this? printf("somevar=''%c''\n", GetSortMethod(anum)); 如果GetSortMethod返回''\ 0''，打印的单引号一定不能用空格分隔 。 - Christopher Benson-Manica |我*应该*知道我在说什么 - 如果我 ataru（at）cyberspace.org |不，我需要知道。火焰欢迎。If GetSortMethod returns ''\0'', the single quotes printed must not beseparated by whitespace.--Christopher Benson-Manica | I *should* know what I''m talking about - if Iataru(at)cyberspace.org | don''t, I need to know. Flames welcome.在文章< bv ********** @ chessie.cirr.com>中，在*** @ nospam.cyberspace.org 说...In article <bv**********@chessie.cirr.com>, at***@nospam.cyberspace.orgsays...（我没写这个代码！） const char * GetSortMethod（unsigned int sortMeth） {char /> const char * sortingstring; if（sortMeth == 0）{ sortingstring =" D"; } if（sortMeth == 1）{ sortingstring =" A" ;; } if（sortMeth == - 1）{ sortingstring ="" ;; } 返回（sortingstring）; } 1）返回一个字符串的自动指针文字是一个很大的错误，对吗？ 2）如果sortMeth不是-1,0或1，那么返回的值比以前更不确定... 3）sortMeth是* never * -1，因为它是未签名的...... 4）以下更正怎么样？ char Ge tSortMethod（int sortMeth） {返回（！sortMeth？ ''D''：sortMeth == 1？ ''A''：0）; } 5）不，开个玩意儿！说真的这次： char GetSortMethod（int sortMeth） { if（sortMeth == 0）返回''D''; 如果（sortMeth == 1）返回''A''; 返回''\ 0''; } 6）使用原始函数像这样： printf（" somevar =''％s''\ nn"，GetSortMethod（anum））; / * int anum; * / 这是使用正确版本的最佳方式吗？ char meth [2]; meth [0] = GetSortMethod（anum）; meth [1] =''\'''; printf（" somevar =''％s''\ n，meth）; < OT> 7）当由C ++编译器（即Borland C ++ Builder 4.0）编译时，原始函数实现是否更可行？ 8）是有任何可以想象的情况，原来的功能实施可能是正确的？即使它真的是一个C ++ 类成员函数？< / OT> 9）任何其他建议或挑剔？ (I did not write this code!) const char *GetSortMethod( unsigned int sortMeth ) { const char *sortingstring; if( sortMeth == 0 ) { sortingstring = "D"; } if( sortMeth == 1 ) { sortingstring = "A"; } if( sortMeth == -1 ) { sortingstring = ""; } return( sortingstring ); } 1) Returning an automatic pointer to a string literal is a big mistake, correct? 2) If sortMeth isn''t -1, 0, or 1, the value returned i even less determinate than it was before... 3) sortMeth is *never* -1, since it''s unsigned... 4) How about the following correction? char GetSortMethod( int sortMeth ) { return( !sortMeth ? ''D'' : sortMeth == 1 ? ''A'' : 0 ); } 5) No, just kidding! Seriously this time: char GetSortMethod( int sortMeth ) { if( sortMeth == 0 ) return ''D''; if( sortMeth == 1 ) return ''A''; return ''\0''; } 6) The original function was used like this: printf( "somevar=''%s''\n", GetSortMethod(anum) ); /* int anum; */ Is this the best way to use the correct version? char meth[2]; meth[0]=GetSortMethod( anum ); meth[1]=''\0''; printf( "somevar=''%s''\n", meth ); <OT> 7) Is the original function implementation any more viable when compiled by a C++ compiler (namely, Borland C++ Builder 4.0)? 8) Is there any conceivable circumstance where the original function implementation could have been correct? Even if it''s really a C++ class member function? </OT> 9) Any other suggestions or nitpicks? 我讨厌多次退货。我在一些实时代码中使用它们，但是我还是讨厌它们。怎么样： char GetSortMethod（int sortMeth） { char ret_val; ret_val = -1; if（sortMeth == 0） ret_val =''D''; if（sortMeth == 1） ret_val =''A''; return（ret_val）; }I hate multiple returns. I have used them in some realtime code, BUT Istill hate them. How about this :char GetSortMethod( int sortMeth ){char ret_val;ret_val = -1;if( sortMeth == 0 )ret_val = ''D'';if( sortMeth == 1 )ret_val = ''A'';return(ret_val);} 这篇关于一种情况的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！