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日期:2025-04-25
2009年1月29日 - int main() { int days[] = {1,2,3,4,5}; int *ptr = days; printf("%u\n", ... No, you can't. The compiler doesn't know what the pointer is pointing to....
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日期:2025-04-27
2013年11月10日 - int largest(int *list, int highest_index) ... There's no way to do that. This is one good reason (among many) to use vectors instead of arrays. But if you ......
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日期:2025-04-24
2011年12月11日 - int testInt[3]; testInt[0] = 0; testInt[1] = 1; testInt[2] = 1; int* point; point ... The easy answer is no, you cannot. You'll probably want to keep a variable ......
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日期:2025-04-30
2008年10月24日 - if (NULL == (p = calloc(sizeof(struct mystruct) * n,1))) { /* handle error ... No, there is no way to get this information without depending strongly on ......
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日期:2025-04-23
2013年4月2日 - char array[40] = "hello"; // size 40 int length = sizeof array / sizeof ... You can't, I'm afraid. You need to pass the length of the array to anyone who ......
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日期:2025-04-24
2012年2月21日 - i have pointer of array(array which is in memory) .... Please see my article here:How to get size of array from its pointer[^]. Permalink....
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日期:2025-04-29
When sizeof is applied to the name of a static array (not allocated through malloc), the result is the size in bytes of the whole array. This is one of the few exceptions to the rule that the name of an array is converted to a pointer to the first element...
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日期:2025-04-26
Why isn't the size of an array sent as a parameter the same as within main? #include void PrintSize(int p_someArray[10]); int main { int myArray[10]; printf("%d\n", sizeof(myArray)); /* As expected, 40 */ PrintSize(myArray);/* Prints 4, not 40 */ } v...