2015年12月7日 · char *s is a pointer to either a char, or a sequence of char. On a 32bit architecture it will be 4 bytes wide, so sizeof(s) will be 4. A single character is (usually) 1 byte, so sizeof(s[0]) will be 1. Therefore, n will be 0.

2011年9月27日 · const char *str = "Test"; The relevant section of the standard is Appendix C section 1.1: Change: String literals made const. The type of a string literal is changed from “array of char” to “array of const char.” The type of a char16_t string literal is changed from “array of some-integer-type” to “array of const char16_t.”

2022年6月14日 · char* represents the address of the beginning of the contiguous block of memory of char's. You need it as you are not using a single char variable you are addressing a whole array of char's. When accessing this, functions will take the address of the first char and step through the memory. This is possible as arrays use contiguous memory (i.e ...

2014年10月17日 · using array_ref = char (&)[2]; array_ref f(...); The reason for returning a reference to an array rather than an actual array is because of the fact that arrays cannot be returned from functions. It's impossible.

@thom_nic: No, because &a has the type char (*)[6], ie, a pointer to an array of six characters. This isn't the same as char *, so you'll get a warning. However, you can say char *p = a;, because an array reference can decay into a pointer to its first element. I used the longer form here, because it better illustrates what's happening.

2011年4月13日 · If you had longer like CHAR(100) vs VARCHAR(100), which is better in space usage depends on the data you have. In any case, in terms of efficieny of queries, it is always a good idea to have fixed length records when you can afford to have them (the DB can optimize better for fixed length columns).

2010年10月7日 · char() is something which allocates the characters depending on the number specified. It would not adjust back to the length of the data you pass. Hence if you specify 2 in char(2), then your field will always have 2 characters.

First off, as others have said, char[2] is not the same as char*, or at least not usually. char[2] is a size 2 array of char and char* is a pointer to a char. They often get confused because arrays will decay to a pointer to the first element whenever they need to. So this works: char foo[2]; char* bar = foo; But the reverse does not:

2010年10月18日 · When Java was originally designed, it was anticipated that any Unicode character would fit in 2 bytes (16 bits), so char and Character were designed accordingly. In fact, a Unicode character can now require up to 4 bytes. Thus, UTF-16, the internal Java encoding, requires supplementary characters use 2 code units.

2018年12月15日 · for your code snippet, *str holds address to a char and **str holds address to a variable holding address of a char. In another word, pointer to pointer. Whenever, you have *str, only enough memory is allocated to hold a pointer type variable(4 byte on a 32 bit machine). With str[10], memory is already allocated for 10 char.