int *nums=(int *)malloc(sizeof(int)*128);如何给这个赋值

您可以使用循环或者memcpy函数将数组赋值。例如使用循环：

for(int i=0;i<128;i++){
nums[i] = i;
}

或者使用memcpy函数：

int values[128] = {1,2,3,...,128}; //假设有一个数组values存放赋值内容
memcpy(nums, values, sizeof(int)*128); //将values数组拷贝到nums数组中。

相关问题

int *nums=malloc(count *sizeof(int))

This code snippet allocates dynamic memory in the heap for an array of integers of size "count". The pointer "nums" points to the first element of the array.

The "sizeof(int)" operator returns the size of an integer in bytes, and by multiplying it with "count", we get the total amount of memory required for the array.

The "malloc()" function allocates the requested amount of memory and returns a pointer to the first byte of the allocated block. In this case, the "malloc()" function returns a void pointer, which is then typecasted to an integer pointer and assigned to "nums".

优化以下代码 char* p = str; int numCount = 0; while (*p != '\0') { if (*p == ',' || *p == '.') { numCount++; } p++; } numCount++; // 数字个数等于分隔符个数加1 *len = numCount; int* nums = (int*)malloc(sizeof(int) * numCount); char *x = strtok(str,",."); int i = 0; while (x!= NULL) { nums[i++] = atoi(x); x = strtok(NULL, ",."); } return nums; }

char* p = str;
int numCount = 1; // 初始化为1，因为最后一个数字后面没有分隔符
while (*p != '\0') {
if (*p == ',' || *p == '.') {
numCount++;
}
p++;
}
*len = numCount;
int* nums = (int*)malloc(sizeof(int) * numCount);
char *x = strtok(str, ",.");
int i = 0;
while (x != NULL) {
nums[i++] = atoi(x);
x = strtok(NULL, ",.");
}
return nums;

主要的优化有两点：
1. 在计算数字个数时，初始化为1，因为最后一个数字后面没有分隔符。
2. 在使用 `strtok` 函数时，将分隔符字符串改为 ",."，这样就不需要在循环中判断分隔符是 "," 还是 "." 了。