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将一个字符串按照奇偶位分成两个字符串
/*题目:有一个字符串”1a2b3d4z”,;
要求写一个函数实现如下功能,
功能1:把奇数位字符挑选出来,组成一个字符串2,valude 20
功能2:把偶数位字符挑选出来,组成一个字符串1。valude;20分
功能3:把字符串1和字符串2,通过函数参数,传送给main,并打印。
功能4:主函数能测试通过。*/void GetStr1Str2(char *m_sourcestr, char *m_str1, char *m_str2)
{if (NULL == m_sourcestr|| NULL == m_str1 || NULL == m_str2){return;}int i = 0;int j = 0;int t = 0;char *temp = m_sourcestr;while (*m_sourcestr != '\0'){temp = m_sourcestr + t;if (*temp == '\0'){break;}if (t%2 == 0){m_str1[i++] = *temp;}else{m_str2[j++] = *temp;}t++;}m_str1[i] = '\0';m_str2[j] = '\0';
}void main()
{int ret = 0;char s0[] = "1a2b3d4z";printf("s0:%s\n", s0);char s1[20], s2[20];GetStr1Str2(s0, s1, s2);if (ret != 0)printf("失败(%d)\n", ret);else{printf("s1:%s\n", s1);printf("s2:%s\n", s2);}system("pause");
}
结果;