１．下面的代码输出是什么，为什么？
void foo(void)
{
unsigned int a = 6;
int b = -20;
(a+b>6)?puts(">6"):puts("<=6");//puts为打印函数
}
答：输出 >6．就是考察隐式转换．int型变量转化成unsigned int, b成了正数．
2. 运行下面的函数会有什么结果？为什么？//有点象华为的试题
void foo(void)
{
char string[10],str1[10];
int i;
for(i=0;i<10;i++)
{
str1[i] = 'a';
}
strcpy(string, str1);
printf("%s",string);
}
答：首先搞清strcpy函数的实现方法：
char * strcpy(char * strDest,const char * strSrc)
{
if ((strDest == NULL) || (strSrc == NULL))
throw "Invalid argument(s)";
char * strDestCopy = strDest;
while ((*strDest++ = *strSrc++) != '/0');
return strDestCopy;
}
由于str1末尾没有‘/0’结束标志，所以strcpy不知道拷贝到何时结束．
printf函数，对于输出char* 类型，顺序打印字符串中的字符直到遇到空字符（＇＼０＇）或已打印了由精度指定的字符数为止．
3. Implement a string class in C++ with basic functionality like comparison, concatenation, input and output. Please also provide some test cases and using scenarios (sample code of using this class).
Please do not use MFC, STL and other libraries in your implementation.
我的实现方案如下,这道题真地对c++的主要特性都进行了较好地考察.
String.h:
#ifndef STRING_H
#define STRING_H
#include <iostream>
using namespace std;
class String{
public:
String();
String(int n,char c);
String(const char* source);
String(const String& s);
//String& operator=(char* s);
String& operator=(const String& s);
~String();
char& operator[](int i){return a[i];}
const char& operator[](int i) const {return a[i];}//对常量的索引.
String& operator+=(const String& s);
int length();
friend istream& operator>>(istream& is, String& s);//搞清为什么将>>设置为友元函数的原因.
//friend bool operator< (const String& left,const String& right);
friend bool operator> (const String& left, const String& right);//下面三个运算符都没必要设成友元函数,这里是为了简单.
friend bool operator== (const String& left, const String& right);
friend bool operator!= (const String& left, const String& right);
private:
char* a;
int size;
};
#endif
String.cpp:
#include "String.h"
#include <cstring>
#include <cstdlib>
String::String(){
a = new char[1];
a[0] = '/0';
size = 0;
}
String::String(int n,char c){
a = new char[n + 1];
memset(a,c,n);
a[n] = '/0';
size = n;
}
String::String(const char* source){
if(source == NULL){
a = new char[1];
a[0] = '/0';
size = 0;
}
else
{ size = strlen(source);
a = new char[size + 1];
strcpy(a,source);
}
}
String::String(const String& s){
size = strlen(s.a);//可以访问私有变量.
a = new char[size + 1];
//if(a == NULL)
strcpy(a,s.a);
}
String& String::operator=(const String& s){
if(this == &s)
return *this;
else
{
delete[] a;
size = strlen(s.a);
a = new char[size + 1];
strcpy(a,s.a);
return *this;
}
}
String::~String(){
delete[] a;//
}
String& String::operator+=(const String& s){
int j = strlen(a);
int size = j + strlen(s.a);
char* tmp = new char[size+1];
strcpy(tmp,a);
strcpy(tmp+j,s.a);
delete[] a;
a = tmp;
return *this;
}
int String::length(){
return strlen(a);
}
main.cpp:
#include <iostream>
#include "String.h"
using namespace std;
bool operator==(const String& left, const String& right)
{
int a = strcmp(left.a,right.a);
if(a == 0)
return true;
else
return false;
}
bool operator!=(const String& left, const String& right)
{
return !(left == right);
}
ostream& operator<<(ostream& os,String& s){
int length = s.length();
for(int i = 0;i < length;i++)
//os << s.a[i];这么不行,私有变量.
os << s[i];
return os;
}
String operator+(const String& a,const String& b){
String temp;
temp = a;
temp += b;
return temp;
}
bool operator<(const String& left,const String& right){
int j = 0;
while((left[j] != '/0') && (right[j] != '/0')){
if(left[j] < right[j])
return true;
else
{
if(left[j] == right[j]){
j++;
continue;
}
else
return false;
}
}
if((left[j] == '/0') && (right[j] != '/0'))
return true;
else
return false;
}
bool operator>(const String& left, const String& right)
{ int a = strcmp(left.a,right.a);
if(a > 0)
return true;
else
return false;
}
istream& operator>>(istream& is, String& s){
delete[] s.a;
s.a = new char[20];
int m = 20;
char c;
int i = 0;
while (is.get(c) && isspace(c));
if (is) {
do {s.a[i] = c;
i++;
/*if(i >= 20){
cout << "Input too much characters!" << endl;
exit(-1);
}*/
if(i == m - 1 ){
s.a[i] = '/0';
char* b = new char[m];
strcpy(b,s.a);
m = m * 2;
s.a = new char[m];
strcpy(s.a,b);
delete[] b;
}
}
while (is.get(c) && !isspace(c));
//如果读到空白,将其放回.
if (is)
is.unget();
}
s.size = i;
s.a[i] = '/0';
return is;
}
int main(){
String a = "abcd";
String b = "www";
//String c(6,b);这么写不对.
String c(6,'l');
String d;
String e = a;//abcd
String f;
cin >> f;//需要输入...
String g;
g = a + b;//abcdwww
if(a < b)
cout << "a < b" << endl;
else
cout << "a >= b" << endl;
if(e == a)
cout << "e == a" << endl;
else
cout << "e != a" << endl;
b += a;
cout << a << endl;
cout << b << endl;
cout << c << endl;
cout << d << endl;
cout << e << endl;
cout << f << endl;
cout << g << endl;
cout << g[0] << endl;
return 0;
}
4. Implement a single-direction linked list sorting algorithm. Please first define the data structure of linked list and then implement the sorting algorithm.
5.编写一个函数，返回两个字符串的最大公串！例如，“adbccadebbca”和“edabccadece”,返回“ccade”
1．设计函数 int atoi(char *s)。
int atoi(const char *nptr);
函数说明
atoi()会扫描参数nptr字符串，跳过前面的空格字符，直到遇上数字或正负符号才开始做转换，而再 遇到非数字或字符串结束时('/0')才结束转换，并将结果返回。
返回值 返回转换后的整型数。
#include <stdio.h>
#include <ctype.h>
int myAtoi(const char* s){
int result = 0;
int flag = 1;
int i = 0;
while(isspace(s[i]))
i++;
if(s[i] == '-'){
flag = -1;
i++;
}
if(s[i] == '+')
i++;
while(s[i] != '/0'){
if((s[i] > '9') || (s[i] < '0'))
break;
int j = s[i] - '0';
result = 10 * result + j;
i++;
}
result = result * flag;
return result;
}
int main(){
char* a = " -1234def";
char* b = "+1234";
int i = myAtoi(a);
int j = myAtoi(b);
printf("%d /n",i);
printf("%d",j);
return 0;
}
2．int i=(j=4,k=8,l=16,m=32); printf(“%d”, i); 输出是多少？
7．下列哪两个是等同的
int b;
A const int* a = &b;
B const* int a = &b;
C const int* const a = &b;
D int const* const a = &b;
8．内联函数在编译时是否做参数类型检查？
void g(base & b){
b.play;
}
void main(){
son s;
g(s);
return;
}
1．完成下列程序
*
*.*.
*..*..*..
*...*...*...*...
*....*....*....*....*....
*.....*.....*.....*.....*.....*.....
*......*......*......*......*......*......*......
*.......*.......*.......*.......*.......*.......*.......*.......
#include <stdio.h>
#define N 8
int main()
{
int i;
int j;
int k;
---------------------------------------------------------
| |
| |
| |
---------------------------------------------------------
return 0;
}
2．完成程序，实现对数组的降序排序
#include <stdio.h>
void sort( );
int main()
{
int array[]={45，56，76，234，1，34，23，2，3}； //数字任//意给出
sort( );
return 0;
}
void sort( )
{
____________________________________
| |
| |
|-----------------------------------------------------|
}
3．费波那其数列，1，1，2，3，5……编写程序求第十项。可以用递归，也可以用其他方法，但要说明你选择的理由。
#include <stdio.h>
int Pheponatch(int);
int main()
{
printf("The 10th is %d",Pheponatch(10));
return 0;
}
int Pheponatch(int N)
{
--------------------------------
| |
| |
--------------------------------
}
4．下列程序运行时会崩溃，请找出错误并改正，并且说明原因。
#include <stdio.h>
#include <malloc.h>
typedef struct{
TNode* left;
TNode* right;
int value;
} TNode;
TNode* root=NULL;
void append(int N);
int main()
{
append(63);
append(45);
append(32);
append(77);
append(96);
append(21);
append(17); // Again, 数字任意给出
}
void append(int N)
{
TNode* NewNode=(TNode *)malloc(sizeof(TNode));
NewNode->value=N;
if(root==NULL)
{
root=NewNode;
return;
}
else
{
TNode* temp;
temp=root;
while((N>=temp.value && temp.left!=NULL) || (N<temp. value && temp. right!=NULL
))
{
while(N>=temp.value && temp.left!=NULL)
temp=temp.left;
while(N<temp.value && temp.right!=NULL)
temp=temp.right;
}
if(N>=temp.value)
temp.left=NewNode;
else
temp.right=NewNode;
return;
}
}
2写出下列程序在X86上的运行结果。
struct mybitfields
{
unsigned short a : 4;
unsigned short b : 5;
unsigned short c : 7;
}test
void main(void)
{
int i;
test.a=2;
test.b=3;
test.c=0;
i=*((short *)&test);
printf("%d/n",i);
}
3写出下列程序的运行结果。
unsigned int i=3;
cout<<i * -1;
4写出下列程序所有可能的运行结果。
int a;
int b;
int c;
void F1()
{
b=a*2;
a=b;
}
void F2()
{
c=a+1;
a=c;
}
main()
{
a=5;
//Start F1,F2 in parallel
F1(); F2();
printf("a=%d/n",a);
}
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