C++ 扫描字符串每个字符的 ASCII 值
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Scanning ASCII value of each character of a string
提问by Kamal Kafkaesque
Is there anyway , if I enter any string , then I want to scan ASCII value of each character inside that string , if I enter "john" then I should get 4 variables getting ASCII value of each character, in C or C++
无论如何，如果我输入任何字符串，那么我想扫描该字符串中每个字符的 ASCII 值，如果我输入“john”，那么我应该得到 4 个变量来获取每个字符的 ASCII 值，在 C 或 C++ 中
采纳答案by Omkant
yeah it's very easy ..just a demo
是的，这很容易..只是一个演示
int main()
{
char *s="hello";
while(*s!='char s[] = "john";
')
{
printf("%c --> %d\n",*s,*s);
s++;
}
return 0;
}
But make sure your machine is supporting the ASCII value format.
In C every char has one integral value associted with it called ASCII.
Using %dformat specifier you can directly print the ASCII of any char as above.
但请确保您的机器支持 ASCII 值格式。在 C 中，每个字符都有一个与之相关的整数值，称为 ASCII。使用%d格式说明符，您可以直接打印上述任何字符的 ASCII。
NOTE: It's better to get good book and practice this kind of program yourself.
注意：最好自己找一本好书并练习这种程序。
回答by Mike Seymour
Given a string in C:
给定一个 C 中的字符串：
std::string s = "john";
or in C++:
或在 C++ 中：
printf("%d", s[0]); // in C
std::cout << static_cast<int>(s[0]); // in C++
s[0]gives the numeric value of the first character, s[1]the second an so on.
s[0]给出第一个字符的数值，s[1]第二个以此类推。
If your computer uses an ASCII representation of characters (which it does, unless it's something very unusual), then these values are the ASCII codes. You can display these values numerically:
如果您的计算机使用字符的 ASCII 表示（它确实如此，除非它非常不寻常），那么这些值就是 ASCII 代码。您可以数字显示这些值：
for (char const * p = s; *p; ++p) {
// Do something with the character value *p
}
Being an integer type (char), you can also assign these values to variables and perform arithmetic on them, if that's what you want.
作为整数类型 ( char)，您还可以将这些值分配给变量并对它们执行算术运算，如果这是您想要的。
I'm not quite sure what you mean by "scan". If you're asking how to iterate over the string to process each character in turn, then in C it's:
我不太确定你所说的“扫描”是什么意思。如果您问如何遍历字符串以依次处理每个字符，那么在 C 中它是：
for (char c : s) {
// Do something with the character value c
}
and in (modern) C++:
在（现代）C++ 中：
char s[SOME_SIZE_YOU_HOPE_IS_LARGE_ENOUGH];
fgets(s, sizeof s, stdin);
If you're asking how to read the string as a line of input from the terminal, then in C it's
如果你问如何从终端读取字符串作为一行输入，那么在 C 中它是
std::string s;
std::cin >> s; // if you want a single word
std::getline(std::cin, s); // if you want a whole line
and in C++ it's
在 C++ 中它是
char c = 'b';
int i = c; //i contains ascii value of char 'b'
If you mean something else by "scan", then please clarify.
如果您的“扫描”是指其他意思，请澄清。
回答by john
There is no way to turn a string of length 'x' into x variables. In C or C++ you can only declare a fixed number of variables. But probably you don't need to do what you are saying. Perhaps you just need an array, or most likely you just need a better way to solve whatever problem you are trying to solve. If you explain what the problem is in the first place, then I'm sure a better way can be explained.
无法将长度为 'x' 的字符串转换为 x 变量。在 C 或 C++ 中，您只能声明固定数量的变量。但可能你不需要做你所说的。也许您只需要一个数组，或者很可能您只需要一种更好的方法来解决您要解决的任何问题。如果您首先解释问题是什么，那么我相信可以解释更好的方法。
回答by Philipp
You can simply get the ascii value of a char by casting it to type int:
您可以通过将字符转换为 int 类型来简单地获取字符的 ascii 值：
#include <iostream>
#include <string>
using std::string;
using std::cout;
using std::endl;
int main()
{
string text = "John";
for (int i = 0; i < text.size(); i++)
{
cout << (int)text[i] << endl; //prints corresponding ascii values (one per line)
}
}
Thus, in your example the code to get the ascii values of a string would look something like this:
因此，在您的示例中，获取字符串的 ascii 值的代码如下所示：
char c = (char)74 // c contains 'J'
To get the corresponding char from an integer representing an entry in the ascii table, you just have to cast the int back to char again:
要从表示 ascii 表中条目的整数中获取相应的 char，您只需将 int 再次转换为 char：
#include <stdio.h>
#include <string.h>
#include <malloc.h>
int main(){
char s[]="abc";
int cnt=0;
while(1){
if(s[cnt++]==NULL)break;
}
int *a=(int *)malloc(sizeof(int)*cnt);
for(int i=0;i<cnt;i++)a[i]=s[i];
for(int i=0;i<cnt-1;i++)printf("%d\n",a[i]);
return 0;
}
The code given above was written in C++ but it basically works the same way in C (and many other languages as well I guess)
上面给出的代码是用 C++ 编写的，但它在 C 中的工作方式基本相同（我猜还有许多其他语言）
回答by Nitin Khanna
Ya,I think there are some more better solutions are also available but this one also be helpful. In C
是的，我认为还有一些更好的解决方案可用，但这个也有帮助。 在 C
#include <iostream>
#include <string>
using namespace std;
int main(){
string s="abc";
//int *a=new int[s.length()];
//for(int i=0;i<s.length();i++)a[i]=s[i];
for(int i=0;i<s.length();i++)
cout<<(int)s[i]<<endl;
return 0;
}
In C++
在 C++ 中
##代码##I hope this one will be helpful..
我希望这个会有所帮助..
