STL测试

来源：互联网发布：淘宝运营规划方案编辑：程序博客网时间：2024/04/28 21:19

//输出OK2vector<BYTE>vb1;vector<BYTE>vb2;vb1.push_back(49);vb1.push_back(49);vb1.push_back('/0');vb1.push_back('/0');vb1.push_back(49);vb2.push_back(49);vb2.push_back(49);vb2.push_back('/0');vb2.push_back('/0');vb2.push_back('/0');if( vb1 == vb2 )printf("OK1/n");elseprintf("OK2/n");return;

//输出://255 -1 FF FF//OK1unsigned charc1;charc2;c1=255;c2=(char)255;printf("%d %d %X %X /n", c1, c2, c1, (unsigned char)c2);if( c1 == (unsigned char)c2 )printf("OK1/n");elseprintf("OK2/n");return;

strings1;strings2;s1="12345";//s2="12345";s1[0]=255;s1[1]=49;s1[2]='/0';s1[3]=49;s1[4]=49;s2+=49;s2+=49;s2+='/0';s2+=49;s2+=49;//s2[0]=49;//s2[1]=49;//s2[2]='/0';//s2[3]='/0';//s2[4]=49;//s2.insert( s2.begin(), 50);printf("%d %s/n", s1.size(), s1.data());printf("%d %s/n", s2.size(), s2.data());if( s1 == s2 )printf("OK1/n");elseprintf("OK2/n");return;

BYTEc1;c1=255;if( (BYTE)s1[0] == c1 )printf("OK1 %d %d /n", (BYTE)s1[0], c1);elseprintf("OK2 %d %d /n", s1[0], c1);return;