C++ 随机字符
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C++ random char
提问by juekvwei
I'm trying to generate a random char like this:
我正在尝试生成这样的随机字符：
char* test = "myname" + GenerateRandomChar();
I'm trying to do it without using string. So please don't say use string. Thanks
我试图在不使用字符串的情况下做到这一点。所以请不要说使用字符串。谢谢
回答by pm100
if you are writing c code (which it seems you are) then do
如果你正在编写 c 代码（看起来你是）那么做
char buff[8]; //sizeof("myname") + 1 + 1
sprintf(buff,"myname%c", GenerateRandomChar());
I am assuming that GenerateRandomChar exists and works
我假设 GenerateRandomChar 存在并且有效
回答by Remy Lebeau
Try something like this:
尝试这样的事情：
char[] test = "mynamechar GenerateRandomChar()
{
return (char) (random(127-33) + 33);
}
"; // note char[] instead of char*
test[6] = GenerateRandomChar();
#include <cstdio>
#include <iostream>
char getRandomChar(){
static char c = 'A' + rand()%24;
return c;
}
int main ()
{
srand((unsigned)time(0));
char str[] = "test";
char buffer[sizeof(str)+1];
char rnd = getRandomChar();
sprintf(buffer, "%s%c",str,rnd);
std::cout << buffer << std::endl;
}
回答by Javi V
You can add a sizeof to the buffer so it is char length independent:
您可以将 sizeof 添加到缓冲区，使其与字符长度无关：
"myname" + c
Note that the getRandomChar()function only gives uppercase letters. You will need to it a bit more sofisticated if you want also lowercase.
请注意，该getRandomChar()函数仅提供大写字母。如果你还想要小写，你需要更复杂一些。
This solution assumes you know the size of the initial string in compilation time.
此解决方案假设您知道编译时初始字符串的大小。
回答by Javi V
Your code as it stands is incorrect for two reasons:
由于两个原因，您的代码不正确：
- You're performing a deprecated conversion from
const char*tochar*. It should not compile. - You're performing pointer arithmetic, not concatenation.
- 您正在执行从
const char*到的弃用转换char*。它不应该编译。 - 您正在执行指针算术，而不是串联。
The following expression:
以下表达式：
char *ptr = "myname";
char c = ' '; // ASCII equivalent is 32
int n = (int) c;
char *s = ptr + n; // s now points out of bounds of ptr
where cis some chardoes not perform concatenation, but rather pointer arithmetic. It's similar to:
where cis somechar不执行连接，而是执行指针运算。它类似于：
#include <cstdlib>
#include <cstring>
#include <iostream>
char foo() {
// return a random character here
}
int main() {
const char* myname = "myname";
// enough room to contain new character
// and null terminator
char* buffer = new char[strlen(myname) + 2];
// strcpy includes the null terminator
strcpy(buffer, myname);
// We overwrite myname's null terminator
buffer[strlen(myname)] = foo();
buffer[strlen(myname) + 1] = '##代码##';
std::cout << buffer;
delete[] buffer;
}
If you insist on doing C-style code, then you need to use the standard library functions strcpyand possibly strcat. Example:
如果您坚持使用 C 风格的代码，那么您需要使用标准库函数strcpy，可能还需要使用strcat. 例子：
