C++ char b[] 的初始值设定项过多
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Too many initializers for char b[]
提问by Artiza Ali
It is morse code program.
I am getting error of too many initializers for char b[]. How can I get rid of this error?
这是莫尔斯电码程序。
我收到错误too many initializers for char b[]。我怎样才能摆脱这个错误？
#include<iostream>
using namespace std;
int main(){
char a[72]={'A','a','B','b','C','c','D','d','E','e','F','f','G','g','H','h','I','i','J','j','K','k','L','l','M','m','N','n','O','o','P','p','Q','q','R','r','S','s','T','t','U','u','V','v','W','w','X','x','Y','y','z','Z','0','1','2','3','4','5','6','7','8','9','.',',','?','\'','!','/','(',')','&','@'};
char b[]={".-",".-","-...","-...","-.-.","-.-.","-..","-..",".",".","..-.","..-.","--.","--.","....","....","..","..",".---",".---","-.-","-.-",".-..",".-..","--","--","-.","-.","---","---",".--.",".--.","--.-","--.-",".-.",".-.","...","...","-","-","..-","..-","...-","...-",".--",".--","-..-","-..-","-.--","-.--","--..","--..","-----","-----",".----",".----","..---","..---","...--","...--","....-","....-",".....",".....","-....","-....","--...","--...","---..","---..","----.","----.",".-.-.-",".-.-.-","--..--","--..--","..--..","..--..",".----.",".----.","-.-.-","-.-.--","-..-.","-..-.","-.--.","-.--.","-.--.-","-.--.-",".-...",".-..."};
char c[40];
cout<<"Enter code ";
cin.getline(c,40);
for(int i=0;i<1;i++){
for(int j=0;j<72;j++){
if(b[j]==c[i]){
cout<<a[j];
}
}
}
return 0;
}
回答by David Heffernan
You have said that bis an array of char. But you are supplying string literals rather than individual characters. It's impossible to know what you really mean to do. Perhaps you actually want bto be an array of strings:
你说过这b是一个char. 但是您提供的是字符串文字而不是单个字符。不可能知道你真正想做什么。也许你真的想b成为一个字符串数组：
const char* b[] = {".-", ".-", "-...", "-...", ...};
回答by bblincoe
A chararray cannot contain strings! You should initialize it with individual chars!
一个char数组不能包含字符串！你应该用单独的chars初始化它！
For example: char b[] = { 'I', 'O', 'U' };
例如： char b[] = { 'I', 'O', 'U' };
When you desire string literals, you can use the following:
当您需要字符串文字时，您可以使用以下内容：
const char* b[] = { "II", "OO", "UU" };
回答by Eric Fortin
Either bshould be an array of const char*or you should put char in the initializer list instead of string literals.
要么b应该是一个数组，const char*要么你应该把 char 放在初始化列表中而不是字符串文字中。
const char* b[]={".-",".-", ...}
or
或者
char b[] = {'a', 'b', ...};
回答by Mad Physicist
You are not trying to store chars in b, but strings. Declare bas const char *b[72] = ...instead.
您不是要char在 s 中存储s b，而是在字符串中存储。声明b作为const char *b[72] = ...代替。
回答by SHIWAM KUMAR PRASAD
write
写
static const char *a[72] =
{
'A', 'a', 'B', 'b', 'C', 'c', 'D', 'd', 'E', 'e', 'F', 'f', 'G', 'g',
'H', 'h', 'I', 'i', 'J', 'j', 'K', 'k', 'L', 'l', 'M', 'm', 'N', 'n',
'O', 'o', 'P', 'p', 'Q', 'q', 'R', 'r', 'S', 's', 'T', 't', 'U', 'u',
'V', 'v', 'W', 'w', 'X', 'x', 'Y', 'y', 'z', 'Z', '0', '1', '2', '3',
'4', '5', '6', '7', '8', '9', '.', ',', '?', '\'', '!', '/', '(', ')',
'&', '@'
};
static const char *b[] =
{
".-", ".-", "-...", "-...", "-.-.", "-.-.", "-..", "-..", ".", ".",
"..-.", "..-.", "--.", "--.", "....", "....", "..", "..", ".---", ".---",
"-.-", "-.-", ".-..", ".-..", "--", "--", "-.", "-.", "---", "---",
".--.", ".--.", "--.-", "--.-", ".-.", ".-.", "...", "...", "-", "-",
"..-", "..-", "...-", "...-", ".--", ".--", "-..-", "-..-", "-.--", "-.--",
"--..", "--..", "-----", "-----", ".----", ".----", "..---", "..---", "...--",
"...--", "....-", "....-", ".....", ".....", "-....", "-....",
"--...", "--...", "---..", "---..", "----.", "----.", ".-.-.-",
".-.-.-", "--..--", "--..--", "..--..", "..--..", ".----.", ".----.",
"-.-.-", "-.-.--", "-..-.", "-..-.", "-.--.", "-.--.", "-.--.-", "-.--.-",
".-...", ".-..."
};
I think it will help you...
我想它会帮助你...
