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copied!<p>Of course your macro can be independent of type specifier, this is one of the things that makes macros so powerful. You can either:</p> <ol> <li>Pass the variable name and the type and let the macro deal with the type.</li> <li>Pass the variable name <em>without</em> the type and assume that the macro performs legal operations on that variable, regardless of the type.</li> </ol> <p>Your actual problem is that in the first form you expect your macro to return an expression, but inside it you declare variables and what-not. I believe that because of that, you put the curly braces in the macro, and now it expands to: <code>({ bla bla })</code>, which would give you the following error message in ISO C99:</p> <blockquote> <p><code>ISO C forbids braced-groups within expressions</code></p> </blockquote> <p>Unfortunately, in ISO C99 you can't both have an expression AND declare temporary variables inside it. This is what functions are made for.</p> <p>So either use the second form that you described, that is pass the "result" variable to the macro:</p> <pre><code>#define min(type_, x_, y_, result_) \ do { \ type_ __tmp1 = (x_); \ type_ __tmp2 = (y_); \ result_ = (__tmp1 < __tmp2 ? __tmp1 : __tmp2); \ } while(0) </code></pre> <p>or use an inline function instead.</p> <p>P.S:<br/> Also, <a href="https://stackoverflow.com/questions/154136/why-are-there-sometimes-meaningless-do-while-and-if-else-statements-in-c-c-mac">this</a> and <a href="https://stackoverflow.com/questions/1067226/c-multi-line-macro-do-while0-vs-scope-block">this</a> are worth reading for good practices when writing C macros.</p>

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