英文:
Is there any difference between these two?
问题
我不知道这是否可能，但我感到困惑。
它们是否相同？
我知道在第一种情况下，我们动态分配了20个int类型元素的内存。
int *p;
p=(int *) malloc(20*sizeof(int));
和
p=(int *) malloc(sizeof(int)*20);
英文:
I donno if this is possible or not but am confused.
Are they both same?
I know that in the first case we are allocating memory dynamically for 20 elements of type int.
int *p;
p=(int *) malloc(20*sizeof(int));
and
p=(int *) malloc(sizeof(int)*20);
答案1
得分: 2
请注意，sizeof 操作符返回类型：size_t。
在int * size_t (20*sizeof(int)) 与 size_t * int (sizeof(int) * 20) 的情况下：乘积与较窄的类型首先扩展到另一个类型，而在C中乘法是_可交换的_：a*b 等于 b*a。
情况会发生变化，当涉及 int * int * size_t 与 size_t * int * int 时，因为在C中，乘法不是_结合的_。首先会进行 int * int（产生 int 乘积），然后再进行 int * size_t。对于某些特定值，第一次乘法可能会溢出，而 size_t * int * int 则不会。
当有多个对象需要相乘时，最好确保最宽的乘法首先进行，例如 size_t * int * int。
> 这两者之间有什么区别吗？
只有两个对象要相乘时，可以根据自己的喜好编写代码。
我喜欢以可能更宽的类型开始。
由于不需要进行强制类型转换，并且根据对象的大小来编写代码更容易进行正确的编码、审查和维护，请考虑以下方式：
p = malloc(sizeof p[0] * 20);
if (p == NULL) {
TBD(); // 处理内存不足情况
}
英文:
Recall that sizeof operator returns type: size_t.
In the case of int * size_t (20*sizeof(int)) versus size_t * int (sizeof(int) * 20): the product is the same as the narrower type is widened to the other first and multiplication in C is commutative: a*b equals b*a.
The situation changes with int * int * size_t versus size_t * int * int as in C, multiplication is not associative. The first multiples int * int (with an int product), then does int * size_t. With select values, the first multiplication may overflow whereas size_t * int * int does not.
When more than 2 objects may be multiplied, best to make certain the widest multiplication happens first and subsequently: example size_t * int * int.
> Is there any difference between these two?
With only 2 objects to multiply, code in whatever way you like.
I like to lead with the likely wider type.
Since the cast is not needed and sizing to the object is easier to code right, review and maintain, consider:
p = malloc(sizeof p[0] * 20);
if (p == NULL) {
TBD(); // Handle out-of-memory
}
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