对redis中单元测试框架的简单修改

Emma ·

更新时间:2024-11-14

· 782 次阅读

　　近在看redis的sds，发现原来redis提供了一个简单的测试框架，觉得挺有新意的。之前写代码，都是单独写所有的单元测试代码，感觉很麻烦，这个还不错，感觉修改一下会更加方便使用。　　测试宏定义如下：

/* This is a really minimal testing framework for C. * * Example: * * test_cond("Check if 1 == 1"， 1==1) * test_cond("Check if 5 > 10"， 5 > 10) * test_report() * * ---------------------------------------------------------------------------- * * Copyright (c) 2010-2012， Salvatore Sanfilippo <antirez at gmail dot com> * All rights reserved. * * Redistribution and use in source and binary forms， with or without * modification， are permitted provided that the following conditions are met: * * * Redistributions of source code must retain the above copyright notice， * this list of conditions and the following disclaimer. * * Redistributions in binary form must reproduce the above copyright * notice， this list of conditions and the following disclaimer in the * documentation and/or other materials provided with the distribution. * * Neither the name of Redis nor the names of its contributors may be used * to endorse or promote products derived from this software without * specific prior written permission. * * THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" * AND ANY EXPRESS OR IMPLIED WARRANTIES， INCLUDING， BUT NOT LIMITED TO， THE * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE * ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE * LIABLE FOR ANY DIRECT， INDIRECT， INCIDENTAL， SPECIAL， EXEMPLARY， OR * CONSEQUENTIAL DAMAGES (INCLUDING， BUT NOT LIMITED TO， PROCUREMENT OF * SUBSTITUTE GOODS OR SERVICES; LOSS OF USE， DATA， OR PROFITS; OR BUSINESS * INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY， WHETHER IN * CONTRACT， STRICT LIABILITY， OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) * ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE， EVEN IF ADVISED OF THE * POSSIBILITY OF SUCH DAMAGE. */ #ifndef __TESTHELP_H #define __TESTHELP_H int __failed_tests = 0; int __test_num = 0; #define test_cond(descr，_c) do { __test_num++; printf("%d - %s: "， __test_num， descr); if(_c) printf("PASSED "); else {printf("FAILED "); __failed_tests++; printf("************************ WARNING *************************** [*failed*] We have failed tests here : %s :%d "，__FILE__， __LINE__); printf("are you continue[y]:"); if(tolower(getchar())=='n'){ test_report(); printf("************************************************************ "); exit(1);} else{ printf("************************************************************ "); }} } while(0); #define test_report() do { printf("------------------------------------------------------------ "); printf("%d tests， %d passed， %d failed "， __test_num， __test_num-__failed_tests， __failed_tests); printf("------------------------------------------------------------ "); } while(0); #endif

　　写一个简单的测试来展现一下：

#include<stdio.h> #include<stdlib.h> #include"testhelp.h" int fun1(int a，int b){ return a+b; } int fun2(int a，int b){ return a-b; } int fun3(int a，int b){ return a*b; } int fun4(int a，int b){ return a/b; } int main(){ test_cond("jiafa"， fun1(1，2)==3); test_cond("jianfa"，fun2(2，1)==2); test_cond("chengfa"，fun3(2，1)==2); test_cond("chufa"，fun4(2，1)==2); test_report(); }

　　可以得到如下的测试结果：