我不确定这样的方法是否适合您?我认为这也应该适用于 C++11,如果需要,可以适用于 std::span,我相信。它只分配一次内存,然后将提供给concat 的每个内存块复制到这个分配的块中,无需任何重新分配。
#include <cassert>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <utility>
template <typename T>
size_t concatSize(T* arr, size_t len) {
return len;
}
template <typename T, typename... Args>
size_t concatSize(T* arr, size_t len, Args... args) {
return len + concatSize(args...);
}
template <typename T>
void copyMem(T* res, T* arr, size_t len) {
memcpy(res, arr, len);
}
template <typename T, typename... Args>
void copyMem(T* res, T* arr, size_t len, Args... args) {
memcpy(res, arr, len);
copyMem((T*)((char*)res + len), args...);
}
template <typename T, typename... Args>
std::pair<T*, size_t> concat(T* arr, size_t len, Args... args) {
static_assert(sizeof... (Args)%2 == 0, "Not even number of arguments!");
auto size = concatSize(arr, len, args...);
T* result = (T*)malloc(size);
copyMem(result, arr, len, args...);
return std::make_pair(result, size);
}
int main() {
{
char a[] = {'a', 'b', 'c'};
char b[] = {'d', 'e', 'f'};
char c[] = {'g', 'h', 'i'};
auto result = concat(a, sizeof a, b, sizeof b, c, sizeof c);
//Don't use regular printf %s since it's not a C-string!
for(int i = 0; i < result.second; ++i) {
std::printf("%c ", result.first[i]);
}
std::printf("\n");
free(result.first);
}
{
int a[] = {'1', '2', '3'};
int b[] = {'4', '5', '6'};
int c[] = {'7', '8', '9'};
auto result = concat(a, sizeof( a), b, sizeof( b), c, sizeof( c));
char* x = (char*)result.first;
for(int i = 0; i < result.second; ++i) {
std::printf("%c ", x[i]);
}
std::printf("\n");
free(result.first);
}
}
Output:
a b c d e f g h i
1 2 3 4 5 6 7 8 9