Question

我编写了以下代码来计算给定char数组中字符的出现次数

#include <iostream>
#include <string>
using namespace std;
int main(){

char f[]={'a','a','b','c','c','d','d','f','e','e'};
int n=sizeof(f)/sizeof(char);
char max=f[0];
for (int i=1;i<n;i++)
{
    if(f[i]>max)
    {
        max=f[i];
    }
}

int m=(int)(max)+1;
int  *b=new int [m];

for (int i=0;i<n;i++)
{
    b[(int)(f[i])]++;
}

for (int i=0;i<m;i++)
{
    if(b[i]!=0)
    {
    cout<<(char)(i)<<" occurs  "<<b[i]<< "times "<<endl;   
    }      
}

return 0;
}

但它显示非常糟糕的结果

 occurs  -842150451times
☺ occurs  -842150451times
☻ occurs  -842150451times
♥ occurs  -842150451times
♦ occurs  -842150451times
♣ occurs  -842150451times
♠ occurs  -842150451times
 occurs  -842150451times
 occurs  -842150451times
         occurs  -84215045

 occurs  -842150451times
♂ occurs  -842150451times
♀ occurs  -842150451times
 occurs  -842150451times
♫ occurs  -842150451times
☼ occurs  -842150451times
► occurs  -842150451times
◄ occurs  -842150451times
↕ occurs  -842150451times
‼ occurs  -842150451times
¶ occurs  -842150451times
§ occurs  -842150451times
▬ occurs  -842150451times
↨ occurs  -842150451times
↑ occurs  -842150451times
↓ occurs  -842150451times
→ occurs  -842150451times
← occurs  -842150451times
∟ occurs  -842150451times
↔ occurs  -842150451times
▲ occurs  -842150451times
▼ occurs  -842150451times
  occurs  -842150451times
! occurs  -842150451times
" occurs  -842150451times
# occurs  -842150451times
$ occurs  -842150451times
% occurs  -842150451times
& occurs  -842150451times
' occurs  -842150451times
( occurs  -842150451times
) occurs  -842150451times
* occurs  -842150451times
+ occurs  -842150451times
, occurs  -842150451times
- occurs  -842150451times
. occurs  -842150451times
/ occurs  -842150451times
0 occurs  -842150451times
1 occurs  -842150451times
2 occurs  -842150451times
3 occurs  -842150451times
4 occurs  -842150451times
5 occurs  -842150451times
6 occurs  -842150451times
7 occurs  -842150451times
8 occurs  -842150451times
9 occurs  -842150451times
: occurs  -842150451times
; occurs  -842150451times
< occurs  -842150451times
= occurs  -842150451times
> occurs  -842150451times
? occurs  -842150451times
@ occurs  -842150451times
A occurs  -842150451times
B occurs  -842150451times
C occurs  -842150451times
D occurs  -842150451times
E occurs  -842150451times
F occurs  -842150451times
G occurs  -842150451times
H occurs  -842150451times
I occurs  -842150451times
J occurs  -842150451times
K occurs  -842150451times
L occurs  -842150451times
M occurs  -842150451times
N occurs  -842150451times
O occurs  -842150451times
P occurs  -842150451times
Q occurs  -842150451times
R occurs  -842150451times
S occurs  -842150451times
T occurs  -842150451times
U occurs  -842150451times
V occurs  -842150451times
W occurs  -842150451times
X occurs  -842150451times
Y occurs  -842150451times
Z occurs  -842150451times
[ occurs  -842150451times
\ occurs  -842150451times
] occurs  -842150451times
^ occurs  -842150451times
_ occurs  -842150451times
` occurs  -842150451times
a occurs  -842150449times
b occurs  -842150450times
c occurs  -842150449times
d occurs  -842150449times
e occurs  -842150449times
f occurs  -842150450times

请帮我解决问题？

Answer 1

你必须将“malloc”ed（new）区域初始化为0，它包含垃圾。

Answer 2

正如其他人所说的那样，我会给你另一种做法。由于有256个ASCII字符，因此只分配了该大小的字符数组。

char mapCount[256];
std::fill(mapCount, mapCount+256, 0); // std::fill is from <algorithm> header

现在，正如您之前所做的那样，只需遍历输入序列并增加相应的mapCount值。即，

++mapCount[(int)inputSequence[i]];

最后输出mapCount索引，其值大于0.将索引转换为char将为您提供实际字符。正如@Kerrek SB建议的那样，有一种与价值关键关联的概念。您可以使用关联容器std :: map来执行此任务。

Answer 3

好吧，你的b数组没有初始化......

我的2美分

Answer 4

分配的缓冲区应初始化为零

int *b=new int [m]; memset(b, 0, sizeof(int) * m);

Answer 5

不确定我是否想要深入研究这段代码，但是既然你把它标记为C ++，我就推荐C ++这样做：

#include <map>
#include <iostream>

std::map<char, unsigned int> histogram;

// Counting
for (char * it = f; it != f + sizeof(f); ++it)
  ++histogram[*it];

// Reporting
for (std::map<char, unsigned int>::const_iterator it = histogram.begin(); it != histogram.end(); ++it)
  std::cout << "Character '" << it->first << "' appears " << it->second << " times.\n";

没有手动内存分配，不会忘记清理，而且它具有相当自我描述性。（如果数组f作为没有大小信息的指针，则必须单独传递大小信息并使用它代替sizeof(f)。）

Answer 6

实际的“错误”是您没有初始化已分配给零的内存，而新[]不会为您执行此操作。

除了使用new []之外，你实际上并没有编写任何C ++，它都是C.为什么不使用vector？这也会将值初始化为0。

//在顶部

#include <vector>

然后

std::vector<int> b[m];

最后为标识符而不是单个字母使用有意义的名称。

Answer 7

这里有很多问题：

你想在这做什么？

char f[]={'a','a','b','c','c','d','d','f','e','e'};
int n=sizeof(f)/sizeof(char);
char max=f[0];
for (int i=1;i<n;i++)
{
    if(f[i]>max)
    {
       max=f[i];
    }
}

完成后，此代码将为max分配数组中最高字符的值，在本例中为e。

在这里？

int m=(int)(max)+1;
//cout<<m;
int  *b=new int [m];

此代码将分配一个大小为m的新整数数组，该数组将为'e' + 1。几乎肯定不是你想要的。您希望这是字符串的长度（n）。

正如其他人已经说过的那样，b将包含未初始化的值，这也会导致一些不必要的行为。

计算char数组中的字符

7 个答案: