C++ 如何将字符附加到 std::string？
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How to append a char to a std::string?
提问by drowneath
The following fails with the error prog.cpp:5:13: error: invalid conversion from ‘char' to ‘const char*'
以下失败并出现错误 prog.cpp:5:13: error: invalid conversion from ‘char' to ‘const char*'
int main()
{
char d = 'd';
std::string y("Hello worl");
y.append(d); // Line 5 - this fails
std::cout << y;
return 0;
}
I also tried, the following, which compiles but behaves randomly at runtime:
我还尝试了以下内容，它可以编译但在运行时行为随机：
int main()
{
char d[1] = { 'd' };
std::string y("Hello worl");
y.append(d);
std::cout << y;
return 0;
}
Sorry for this dumb question, but I've searched around google, what I could see are just "char array to char ptr", "char ptr to char array", etc.
对不起，这个愚蠢的问题，但我在谷歌周围搜索过，我能看到的只是“char array to char ptr”，“char ptr to char array”等。
回答by AraK
y += d;
I would use +=operator instead of named functions.
我会使用+=运算符而不是命名函数。
回答by Ferdinand Beyer
Use push_back():
使用push_back()：
std::string y("Hello worl");
y.push_back('d')
std::cout << y;
回答by Patrice Bernassola
To add a char to a std::string var using the append method, you need to use this overload:
要使用 append 方法将字符添加到 std::string 变量，您需要使用此重载：
std::string::append(size_type _Count, char _Ch)
Edit : Your're right I misunderstood the size_type parameter, displayed in the context help. This is the number of chars to add. So the correct call is
编辑：您说得对，我误解了上下文帮助中显示的 size_type 参数。这是要添加的字符数。所以正确的调用是
s.append(1, d);
not
不是
s.append(sizeof(char), d);
Or the simpliest way :
或者最简单的方法：
s += d;
回答by Brian R. Bondy
In addition to the others mentioned, one of the string constructors take a char and the number of repetitions for that char. So you can use that to append a single char.
除了提到的其他人之外，其中一个字符串构造函数采用一个字符和该字符的重复次数。因此，您可以使用它来附加单个字符。
std::string s = "hell";
s += std::string(1, 'o');
回答by Hugo Zevetel
I test the several propositions by running them into a large loop. I used microsoft visual studio 2015 as compiler and my processor is an i7, 8Hz, 2GHz.
我通过将它们运行到一个大循环中来测试几个命题。我使用 microsoft visual studio 2015 作为编译器，我的处理器是 i7、8Hz、2GHz。
long start = clock();
int a = 0;
//100000000
std::string ret;
for (int i = 0; i < 60000000; i++)
{
ret.append(1, ' ');
//ret += ' ';
//ret.push_back(' ');
//ret.insert(ret.end(), 1, ' ');
//ret.resize(ret.size() + 1, ' ');
}
long stop = clock();
long test = stop - start;
return 0;
According to this test, results are :
根据这个测试，结果是：
operation time(ms) note
------------------------------------------------------------------------
append 66015
+= 67328 1.02 time slower than 'append'
resize 83867 1.27 time slower than 'append'
push_back & insert 90000 more than 1.36 time slower than 'append'
Conclusion
结论
+=seems more understandable, but if you mind about speed, use append
+=似乎更容易理解，但如果您介意速度，请使用 append
回答by Michael Berg
回答by Alex Spencer
the problem with:
问题在于：
std::string y("Hello worl");
y.push_back('d')
std::cout << y;
is that you have to have the 'd' as opposed to using a name of a char, like char d = 'd'; Or am I wrong?
是你必须有 'd' 而不是使用字符的名称，比如 char d = 'd'; 还是我错了？
回答by Mooing Duck
int main()
{
char d = 'd';
std::string y("Hello worl");
y += d;
y.push_back(d);
y.append(1, d); //appending the character 1 time
y.insert(y.end(), 1, d); //appending the character 1 time
y.resize(y.size()+1, d); //appending the character 1 time
y += std::string(1, d); //appending the character 1 time
}
Note that in all of these examples you could have used a character literal directly: y += 'd';.
请注意，在所有这些示例中，您都可以直接使用字符文字：y += 'd';.
Your second example almostwould have worked, for unrelated reasons. char d[1] = { 'd'};didn't work, but char d[2] = { 'd'};(note the array is size two) would have been worked roughly the same as const char* d = "d";, and a string literal canbe appended: y.append(d);.
由于无关的原因，您的第二个示例几乎会起作用。char d[1] = { 'd'};没有用，但char d[2] = { 'd'};（注意数组大小为 2）的工作方式与 大致相同const char* d = "d";，并且可以附加字符串文字：y.append(d);。
回答by 2ndless
Try using the d as pointer y.append(*d)
尝试使用 d 作为指针 y.append(*d)
回答by Charles H
I found a simple way...
I needed to tack a charon to a string that was being built on the fly. I needed a char list;because I was giving the user a choice and using that choice in a switch()statement.
我找到了一个简单的方法......我需要char在正在构建的字符串上钉上一根绳子。我需要一个，char list;因为我给了用户一个选择并在switch()声明中使用该选择。
I simply added another std::string Slist;and set the new string equal to the character, "list" - a, b, c or whatever the end user chooses like this:
我只是添加了另一个std::string Slist;并将新字符串设置为等于字符“列表”-a、b、c 或最终用户选择的任何内容，如下所示：
char list;
std::string cmd, state[], Slist;
Slist = list; //set this string to the chosen char;
cmd = Slist + state[x] + "whatever";
system(cmd.c_str());
Complexity may be cool but simplicity is cooler. IMHO
复杂可能很酷，但简单更酷。恕我直言
