C++：获取数组中字符元素的索引
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C++: Get index of char element in array
提问by ghostmansd
I need to get number of character in array.
我需要获取数组中的字符数。
const char myarray[5] = {'0', 'a', 'e', 'f', 'c'}; // Create array of char
int number=0; // Create variable
number = getposition(myarray, 'f'); // Now number equals to 3
number = getposition(myarray, 'z'); // -1, because array doesn't have this char
My task is easy because array don't have repeating characters (for example, it can't be smth like this: {'a', '1', 'f', 'a'}). How can I do it?
我的任务很简单，因为数组没有重复字符（例如，它不能像这样：{'a', '1', 'f', 'a'}）。我该怎么做？
回答by sehe
A little bit more C++:
多一点 C++：
#include <algorithm>
int getposition(const char *array, size_t size, char c)
{
const char* end = array + size;
const char* match = std::find(array, end, c);
return (end == match)? -1 : (match-array);
}
A lot more C++:
更多的C++：
template <typename T, size_t N>
int getposition(const T (&array)[N], const T c)
{
const T* match = std::find(array, array+N, c);
return (array+N==match)? -1 : std::distance(array, match);
}
Bonus C++11/C++11 update
额外的 C++11/C++11 更新
#include <algorithm>
#include <iterator>
template <typename Range, typename T>
size_t index_of(Range const& range, T const& c) {
using std::begin;
using std::end;
auto b = begin(range), e = end(range);
auto match = std::find(b, e, c);
return (e==match)? -1 : std::distance(b, match);
}
Bonus C++17 update
额外的 C++17 更新
Here, the original question gets direct support in std::string_view:
在这里，原始问题得到了直接支持std::string_view：
#include <string_view>
using namespace std::string_view_literals;
int main() {
return "hello"sv.find('e');
}
回答by Mike Seymour
#include <algorithm>
template <typename T, size_t size>
int getposition(T const (&array)[size], T const & c)
{
T const * found = std::find(&array[0], &array[size], c);
return found == &array[size] ? -1 : found - array;
}
回答by trojanfoe
You need to tell the getposition()method how many elements to search within the array and as the array is initialised at compile time you can use the sizeofdirective:
您需要告诉该getposition()方法要在数组中搜索多少个元素，并且在编译时初始化数组时，您可以使用该sizeof指令：
int number = getposition(myarray, sizeof(myarray), 'f');
...
int getposition(const char *array, size_t size, char c)
{
for (size_t i = 0; i < size; i++)
{
if (array[i] == c)
return (int)i;
}
return -1;
}
回答by Firedragon
int getposition(const char* a, int arr_size, char to_find)
{
int pos = -1;
for(int i = 0; i < arr_size; ++i)
{
if(a[i] == to_find)
{
pos = i;
break;
}
}
return pos;
}
回答by Nim
If this really is a pure decoding exercise - why not re-organize your array... then lookup is constant time - e.g..
如果这真的是一个纯粹的解码练习——为什么不重新组织你的数组......那么查找是恒定的时间——例如。
int lt[128]; // ignoring negative values..
memset(lt, -1, 128); // initialize all to -1
// set the ones you want mappings for..
lt['0'] = 0;
lt['a'] = 1;
lt['e'] = 2;
lt['f'] = 3;
lt['c'] = 4;
so now your look up function is:
所以现在你的查找功能是：
int indexOf(char v) { return lt[v]; }
You'd be hard-pressed to beat that for performance...
你很难在性能上击败它......
回答by m0skit0
You need to pass the array size as well to the function.
您还需要将数组大小传递给函数。
int getposition(const char* array, size_t array_size, char value)
{
int ret = -1;
int i = 0;
bool found = false;
while (i < array_size && !found)
{
found = (array[i++] == value);
}
if (found)
{
ret = i - 1;
}
return ret;
}
