上帝幫你關了一扇門,就會幫你開一扇窗,
所以,我打算幫準備寫考卷的你 / 妳 開一扇窗,
我將答案放在我的blog裡面,
如果你有看到,請背起來,
這樣你期待薪資的地方,
才有可能成真。
1. Please write the following print.
char s[] = "0113256";
char *p = s;
printf("%c", *p++);
printf("%c", *(p++));
printf("%c", (*p)++);
printf("%c", *++p);
printf("%c", *(++p));
printf("%c", ++*p);
printf("%c", ++(*p));
printf("\n");
printf(s);
1.___________
2.___________
2. Please write the following print.
union AA
{
char a[2];
int s;
};
int main()
{
AA aa = { 0 };
aa.a[0] = 12;
aa.a[1] = 1;
printf("%x\n", aa.s);
printf("%d\n", sizeof(aa));
getchar();
return 0;
}
1.___________
2.___________
3.
#define XPROC(X) X*X
int main()
{
int n = XPROC(2 + 3);
cout << n << endl;
}
n = _________
4.Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.
Example 1:
Input: "Let's take LeetCode contest"
Output: "s'teL ekat edoCteeL tsetnoc"
Note: In the string, each word is separated by single space and there will not be any extra space in the string.
class Solution
{
public:
string reverseWords(string s)
{
return s;
}
};
1.
0113234
0123456
2.
printf("%x\n", aa.s); = 10c
printf("%d\n", sizeof(aa)); =4
3.
n = 11
關於這題,因已不是第一次有人詢問,所以稍微講解一下,
題目是
#define XPROC(X) X*X
int n = XPORC(2+3);
然後問 n = 多少,
這部分是要考Macro的概念,其實還蠻簡單的,
就是將Macro展開,就這樣
也就是說,我們可以把這題看成是
int n = 2+3*2+3
這樣結果,就是等於11了
4.
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