数组的长度计数异常 [英] Array length counting anomaly
问题描述
该计数未返回predictable结果。有时候，他们是对的。有时候完全不可思议。谁能告诉我什么是错？
的#include＆LT;＆stdio.h中GT;
INT LEN（INT []）;INT主（INT ARGC，为const char * argv的[]）
{
诠释一个[] = {1,2,3,4,5};
为int * i为一个;
的printf（长度为％d，（LEN（I）））;
返回0;
}INT LEN（INT *一）{
诠释计数= 0;
对于（; * A ='\\ 0';！一++）{
算上++;
}
返回计数;
}
我觉得你的C字符串（字符的阵列）等阵列之间的混淆。它是C字符串终止与空字符（'\\ 0'），但不是所有的数组（甚至字符数组）终止这种方式。
的一般约定是要么存储数组的长度的地方，或在数组的结尾使用一个哨兵值。这个值应该是一个不上去阵内 - 例如'\\ 0'的字符串或 1 正整数数组。
另外，如果你知道 A 是一个int数组（而不是指向一个int数组），那么你可以使用：
为size_t长度= sizeof的（A）/ sizeof的（A [0]）;
所以，你可以这样做：
int类型的[] = {1,2,3,4,5};
为size_t长度= sizeof的（一）/的sizeof（一个[0]）;//在这种情况下，的sizeof（一个[0]）
//相同的sizeof（int）的，因为它是一个int数组。
但你不能做的：
为int * A =的malloc（sizeof的（INT）* 10）;
为size_t长度= sizeof的（一）/的sizeof（一个[0]）; // WRONG！
这是最后一个例子可以编译，但答案是错误的，因为你得到一个指针的大小数组，而不是数组的大小。
请注意，您也不能使用该的sizeof 读取的被传递给函数的数组的大小。这不要紧，你是否声明你的函数 LEN（INT * A）或 LEN（int类型的[]） - A 将是一个指针，因为编译器在功能参数转换阵列是指向自己的第一个元素
The count is returning unpredictable results. Sometimes they are right. Sometimes totally weird. Anyone can tell me what is wrong?
#include <stdio.h>
int len(int[]);
int main (int argc, const char * argv[])
{
int a[]={1,2,3,4,5,6,7,8};
int* i = a;
printf("length is %d",(len(i)));
return 0;
}
int len(int* a){
int count = 0;
for (; *a!='\0'; a++) {
count++;
}
return count;
}
I think you're confused between C strings (arrays of char) and other arrays. It's a convention that C strings are terminated with a null character ('\0'), but not all arrays (even char arrays) are terminated this way.
The general convention is to either store the length of an array somewhere, or to use a sentinel value at the end of the array. This value should be one that won't come up inside the array - eg '\0' in strings, or -1 in an array of positive ints.
Also, if you know that a is an int array (and not a pointer to an int array), then you can use:
size_t length = sizeof(a) / sizeof(a[0]);
So you could do:
int a[] = {1,2,3,4,5,6,7,8};
size_t length = sizeof(a) / sizeof(a[0]);
// In this case, sizeof(a[0])
// is the same as sizeof(int), because it's an int array.
But you can't do:
int *a = malloc(sizeof(int) * 10);
size_t length = sizeof(a) / sizeof(a[0]); // WRONG!
That last example will compile, but the answer will be wrong, because you're getting the size of a pointer to the array rather than the size of the array.
Note that you also can't use this sizeof to read the size of an array that's been passed into a function. It doesn't matter whether you declare your function len(int *a) or len(int a[]) - a will be a pointer, because the compiler converts arrays in function arguments to be a pointer to their first element.
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