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发表于 2015-11-4 10:15:08
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思路
1. 先将一整个字符串拆成多个数字 并存入数组
2. 通过判断数组内元素的个数 判断是否是合理的参数 超过3个则报错, 小于等于3个则根据元素的个数调用不同的函数
like this?
- #include <iterator>
- #include <vector>
- #include <stdio.h>
- #include <stdlib.h>
- using namespace std;
- template<typename T>
- T cale(T n)
- {
- return n*n;
- }
- template<typename T>
- T cale(T n1, T n2)
- {
- return n1 * n2;
- }
- template<typename T>
- T cale(T n1, T n2, T n3)
- {
- return n1 + n2 + n3;
- }
- void splitstr(const char* str, vector<int>* vec)
- {
- const char* iter = str;
- const char* pos = str;
- while (*iter != NULL)
- {
- pos = strchr(iter, ' ');
- if (pos == NULL) // last char
- {
- vec->push_back(atoi(iter));
- return;
- }
- char buf[0xF] = {0};
- memcpy(buf, iter, pos - iter);
- iter = pos+1;
- vec->push_back(atoi(buf));
- }
- }
- int main (void)
- {
- char buf[0xFF] = {0};
- printf("enter the str\n");
- while (gets(buf) != NULL)
- {
- // init vec
- vector<int> vec;
- vec.reserve(0xFF);
- // split str and save the number into the vector
- splitstr(buf, &vec);
- printf("the argument list is\n");
- copy(vec.begin(), vec.end(), ostream_iterator<int>(cout, " "));
- printf("\n");
- switch(vec.size())
- {
- case 1:
- printf("call cale(T n), res = %d\n\n", cale(*vec.begin()));
- break;
- case 2:
- printf("call cale(T n1, T n2), res = %d\n\n", cale(*vec.begin(), *vec.rbegin()));
- break;
- case 3:
- printf("call cale(T n1, T n2, T n3), res = %d\n\n", cale(*vec.begin(), vec[1], *vec.rbegin()));
- break;
-
- default:
- printf("err the arguments numbers\n\n");
- break;
- }
- }
- return 0;
- }
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