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Array in C

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i*p2007-10-26 07:10

1 楼

Could anybody explain why statement 3 works?
char abc[30];
1. strcpy(&(abc[0], "Hello");
2. strcpy(abc ,"Hello");
3. strcpy(&abc ,"Hello");

k*f2007-10-26 07:10

2 楼

works不等于正确，
会有后遗症的

【在 i**p 的大作中提到】

: Could anybody explain why statement 3 works?
: char abc[30];
: 1. strcpy(&(abc[0], "Hello");
: 2. strcpy(abc ,"Hello");
: 3. strcpy(&abc ,"Hello");

t*t2007-10-26 07:10

3 楼

http://c-faq.com/aryptr/aryvsadr.html

【在 i**p 的大作中提到】

: Could anybody explain why statement 3 works?
: char abc[30];
: 1. strcpy(&(abc[0], "Hello");
: 2. strcpy(abc ,"Hello");
: 3. strcpy(&abc ,"Hello");

i*p2007-10-26 07:10

4 楼

So based on the explaniation in this web page, I have following:
char abc[30];
abc reference to a has type ``pointer to char,'' and &abc is ``pointer to
array of 10 chars.''
we should use &abc for a string, but mostly we use abc. Please comment.

【在 t****t 的大作中提到】

: http://c-faq.com/aryptr/aryvsadr.html

b*a2007-10-26 07:10

5 楼

actually &abc and abc point to the same address.
the different is when you set
char (*p1)[30]=&abc;
p1+=1;
char* p2=abc;
p2+=1;
p1+1 and p2+1 will point to different address,
say p2=p1=0x000010, p1+1 will make p1 point to 0x00010+ 1E*sizeof(char);
p2+1 will point to 0x00010+1*sizeof(char)

【在 i**p 的大作中提到】

: So based on the explaniation in this web page, I have following:
: char abc[30];
: abc reference to a has type ``pointer to char,'' and &abc is ``pointer to
: array of 10 chars.''
: we should use &abc for a string, but mostly we use abc. Please comment.

t*t2007-10-26 07:10

6 楼

comment: you are wrong, and please read the whole chapter of FAQ.

【在 i**p 的大作中提到】

k*f2007-10-26 07:10

7 楼

哈哈，你太nice了。

【在 t****t 的大作中提到】

: comment: you are wrong, and please read the whole chapter of FAQ.

t*t2007-10-26 07:10

8 楼

that... really depends on the type of p1 and p2. if p1 and p2 are of the
same type,
your p1+1 and p2+1 will be the same anyway. if you say (&abc)+1 and abc+1
are different, that makes more sense.

【在 b******a 的大作中提到】

: actually &abc and abc point to the same address.
: the different is when you set
: char (*p1)[30]=&abc;
: p1+=1;
: char* p2=abc;
: p2+=1;
: p1+1 and p2+1 will point to different address,
: say p2=p1=0x000010, p1+1 will make p1 point to 0x00010+ 1E*sizeof(char);
: p2+1 will point to 0x00010+1*sizeof(char)

b*a2007-10-26 07:10

9 楼

哈哈。又被你老给找出毛病来了。

【在 t****t 的大作中提到】

: that... really depends on the type of p1 and p2. if p1 and p2 are of the
: same type,
: your p1+1 and p2+1 will be the same anyway. if you say (&abc)+1 and abc+1
: are different, that makes more sense.