“const char *”类型的值不能分配给“char”类型的实体 C OOP
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a value of type "const char *" cannot be assigned to an entity of type "char" C OOP
提问by Luke
I am creating a class to calculate a grade for a user in C++ and I am coming across a simple yet annoying problem. I know what the errors means but I don't understand how to fix it and changing to a string actually fixes the issue but this is not what I want to do.
我正在创建一个类来为 C++ 中的用户计算成绩，但我遇到了一个简单但烦人的问题。我知道错误意味着什么，但我不明白如何修复它并更改为字符串实际上可以解决问题，但这不是我想要做的。
here is the error: const char *" cannot be assigned to an entity of type "char
这是错误：const char *”不能分配给“char”类型的实体
Code
代码
#include <string>
using namespace std;
class Gradecalc
{
public:
Gradecalc()
{
mark = 0;
}
int getmark()
{
return mark;
}
void setmark(int inmark)
{
mark = inmark;
}
void calcgrade()
{
if (mark >=70)
{
grade = "A"; //**ERROR IS HERE**
}
}
char getgrade()
{
return grade;
}
private:
int mark;
char grade; //VARIABLE IS DECLARED HERE
};
回答by dasblinkenlight
C++ has two types of constants consisting of characters - string literals and character literals.
C++ 有两种类型的由字符组成的常量——字符串字面量和字符字面量。
- String literals are enclosed in double quotes, and have type of
const char * - Character literals are enclosed in single quotes, and have type
char.
- 字符串文字用双引号括起来，类型为
const char * - 字符文字用单引号括起来，类型为
char。
String literals allow multiple characters; character literals allow only one character. The two types of literals are not compatible: you need to supply a variable or a constant of a compatible type for the left side of the assignment. Since you declared gradeas a char, you need to change the code to use a character literal, like this:
字符串文字允许多个字符；字符文字只允许一个字符。这两种类型的文字不兼容：您需要为赋值的左侧提供一个兼容类型的变量或常量。由于您声明grade为 a char，因此您需要更改代码以使用字符文字，如下所示：
grade ='A';
回答by kfsone
C and C++ use double quotes to indicate "string literal", which is verydifferent from a "character literal".
C 和 C++ 使用双引号表示“字符串文字”，这与“字符文字”非常不同。
char(which is signed) is a type capable of storing a character representation in the compiler's default character set. On a modern, Western PC, that means ASCII, which is a character set that requires 7-bits, plus one for sign. So, charis generally an 8-bit value or byte.
char（有符号）是一种能够在编译器的默认字符集中存储字符表示的类型。在现代的西方 PC 上，这意味着 ASCII，这是一个需要 7 位的字符集，加上一个符号。所以，char一般是一个8位的值或字节。
A character literal is formed using single quotes, so 'A'evaluates to ASCII code 65. ('A' == 65).
字符文字是使用单引号形成的，因此'A'计算结果为 ASCII 代码 65. ( 'A' == 65)。
On the other hand, "A"causes the compiler to write char(65) + char(0) into a part of the output program and then evaluates the expression "A"to the address of that sequence; thus it evaluates to a pointer to a char sequence, but they're in the program data itself so they are not modifiable, hence const char*.
另一方面，"A"使编译器将 char(65) + char(0) 写入输出程序的一部分，然后将表达式计算"A"为该序列的地址；因此它评估为指向字符序列的指针，但它们在程序数据本身中，因此它们不可修改，因此const char*.
You want
你要
grade = 'A';
回答by Ilya Kobelevskiy
Replace
代替
grade = "A";
by
经过
grade = 'A';
You can only assign char to char, you cannot assign string to single char, and that is what you are trying to do.
您只能将字符分配给字符，不能将字符串分配给单个字符，这就是您要尝试执行的操作。
回答by Billie
Gradeis a charvariable, "A"is a const char*type.
Grade是char变量，"A"是const char*类型。
You cannot assign const char*into charvarible.
您不能分配const char*到charvarible。
double quote means const char*, and single qoute means char.
双引号表示const char*，单引号表示字符。
to fix that, replace:
要解决这个问题，请替换：
grade="A"
等级=“A”
with
和
grade='A'.
等级='A'。
