&符号和C中的scanf? [英] Ampersand(&) and scanf in C?

问题描述

我有以下多项选择题，我不能弄清楚为什么(A)和(C)不正确，请多加解释！下一个问题中唯一的正确答案是(B).

I have the following multiple-choice question and I cannot figure out why (A) and (C) are incorrect, any explanation would be appreciated! The only correct answer in the following question is (B).

以下哪些是scanf的正确用法?
(A)int i=0; scanf("%d", i);
(B)int i=0; int *p=&i; scanf("%d", p);
(C)char *s="1234567"; scanf("%3s", &s);
(D)char c; scanf("%c", c);

Which of the following is a correct usage of scanf?
(A) int i=0; scanf("%d", i);
(B) int i=0; int *p=&i; scanf("%d", p);
(C) char *s="1234567"; scanf("%3s", &s);
(D) char c; scanf("%c", c);

推荐答案

scanf希望在正确的地址存储结果:

scanf wants a correct address where to store the result:

(A) int i=0; scanf("%d", i);

i是按值传递的，没有地址:错误.

i is passed here by value, no address: wrong.

(B) int i=0; int *p=&i; scanf("%d", p);

p是指向int的指针，scanf可以将其结果存储在此处:对

p is a pointer to int, scanf can store its result here: right

(C) char *s="1234567"; scanf("%3s", &s);

&s是指向char *的指针的地址，您不能在其中存储字符串:错误

&s is the address of a pointer to char *, you cannot store a string there: wrong

(D) char c; scanf("%c", c);

c是通过值而不是地址传递的:错误

c is passed by value, not an address: wrong

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