我以前发布过,但是我没有正确格式化或添加我的代码。假设我有一个整数数组x = [1,2,3]。给定一个值i,我想创建一个数组x ^ i,如果i = 3,则数组x ^ i = [1,1,1,2,2,2,3,3,3]。如果i = 5,数组x ^ i = [1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4 ,5,5,5,5,5]。我正在为此动态分配内存。
但是,我对i = 3的代码正在创建一个数组= [1,2,3,1,2,3,1,2,3]。我尝试了许多不同的方法,得到了类似[1,1,1,1,1,1,1,1,1]或[3,3,3,3,3,3,3,3, 3],但永远不会是正确答案。
这是我的代码:
void binary_search(int size_a, int * A, int size_x, int *X, int max_i, int min_i){
int i, j, k, count = 0, max_repeat = 0;
while(min_i <= max_i){
int repeats = (max_i + min_i)/2;
int * temp = realloc(X, size_x * sizeof(int) * repeats);
X = temp;
for(k = 0; k < size_x; ++k){
int idx = size_x - k -1;
temp = &X[idx];
for(j = 0; j < repeats; ++j){
X[idx * repeats + j] = *temp;
}
}
printf("New X: ");
for(i = 0; i < size_x * repeats; i++){
printf("%d ", X[i]);
}
int count = 0;
for(i = 0; i < size_x * repeats; i++){
for(j = 0; j < size_a; j++){
if(A[j] == X[i]){
count++;
i++;
}
}
}
if (count == size_x * repeats){
printf("Low: %d Mid %d High % d Passes\n", min_i, repeats, max_i);
min_i = repeats + 1;
}
else
printf("Low: %d Mid %d High % d Fails\n", min_i, repeats, max_i);
max_i = repeats - 1;
}
}
变量重复表示x ^ i中的值i。
输出是这样的:
Old X: 1 2 3
New X: 1 1 1 2 2 2 3 3 3 Low: 0 Mid 3 High 6 Fails
New X: 1 1 1 Low: 0 Mid 1 High 2 Fails
New X: Low: 0 Mid 0 High 0 Fails
第一次迭代是正确的,但是第二次迭代不应为[1,1,1],而应为[1,2,3]。
我要去哪里错了?
答案 0 :(得分:0)
您在这里:
int misleading_function_names_is_bad_practice(size_t xsize, int x[xsize], size_t i)
{
void * const tmp = realloc(x, xsize * sizeof(*x) * i);
if (tmp == NULL) {
return -__LINE__;
}
x = tmp;
for (size_t k = 0; k < xsize; ++k) {
// index of the last original digit counting down
const size_t idx = xsize - k - 1;
const int tmp = x[idx];
for (size_t l = 0; l < i; ++l) {
// fill from the back
x[idx * i + l] = tmp;
}
}
return 0;
}
可通过onlinegdb获得实时示例。