我在运行以下代码时遇到了一些问题。我得到了:错误C2668:'pow':对重载函数的模棱两可的调用。我尝试使用static_cast手动将参数强制转换为适当的类型,但是我认为我遇到了一些指针错误?
程序应将数字从16转换为10。
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
#include <string.h>
#include <math.h>
//base 16 to base 10
int convert(char *n){
int result = 0;
for (int i = strlen(n) - 1; i >= 0; i--){
if (n[i] >= 'a')
result += (n[i] - 'a' + 10)* pow(16, strlen(n) - i - 1);
else
if (n[i] >= 'A')
result += (n[i] - 'A' + 10)* pow(16, strlen(n) - i - 1);
else
if (n[i] >= '0')
result += (n[i] - '0')* pow(16, strlen(n) - i - 1);
}
return result;
}
void main(void){
char n[10];
printf("Introduceti numarul: "); scanf("%s", n);
printf("Numarul in baza 10 este: %d", convert(n));
_getch();
}
这些都是错误。
1>------ Build started: Project: pr8, Configuration: Debug Win32 ------
1> pr8.cpp
1> error C2668: 'pow' : ambiguous call to overloaded function
1> could be 'long double pow(long double,int) throw()'
1> or 'long double pow(long double,long double) throw()'
1> or 'float pow(float,int) throw()'
1> or 'float pow(float,float) throw()'
1> or 'double pow(double,int) throw()'
1> or 'double pow(double,double)'
1> while trying to match the argument list '(int, size_t)'
1>'-' : pointer can only be subtracted from another pointer
1> error C2668: 'pow' : ambiguous call to overloaded function
1> could be 'long double pow(long double,int) throw()'
1> or 'long double pow(long double,long double) throw()'
1> or 'float pow(float,int) throw()'
1> or 'float pow(float,float) throw()'
1> or 'double pow(double,int) throw()'
1> or 'double pow(double,double)'
1> while trying to match the argument list '(int, size_t)'
1> error C2668: 'pow' : ambiguous call to overloaded function
1> could be 'long double pow(long double,int) throw()'
1> or 'long double pow(long double,long double) throw()'
1> or 'float pow(float,int) throw()'
1> or 'float pow(float,float) throw()'
1> or 'double pow(double,int) throw()'
1> or 'double pow(double,double)'
1> while trying to match the argument list '(int, size_t)'
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========
我怎样才能解决这个问题?谢谢你。
最佳答案
strlen
返回类型在C++中为size_t
。因此,您可以通过强制转换解决歧义:
pow(static_cast<size_t>(16), strlen(n) - i - 1);
也在这里:
result += (n[i] - "A" + 10)
^ this should be 'A'
并且main应该返回
int
而不是void
:int main(void) {
关于c++ - 对重载函数 'pow'的歧义调用,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/24749711/