Question

我是C的新手。我在GWBASIC方面经验丰富。为了学习，我试图编写一个程序，将字符串中的各个字符转换为数值，如下所示：

1 2 3 4 5 6 7 8 9
a b c d e f g h i
j k l m n o p q r
s t u v w x y z

例如，字符串A的用户输入可能是“狗”，所述程序然后将[d] [o] [g]存储为字符串B中的[4] [6] [7]。下面的代码适用于最多四个字符的字符串，但必须有一种更有效的方法。

int main()
{
    char a[0];
    char b[0];
    scanf("%s",a);
    if (a[0] == 'a' || a[0] == 'j' || a[0] == 's') b[0] = '1';
    if (a[0] == 'b' || a[0] == 'k' || a[0] == 't') b[0] = '2';
    if (a[0] == 'c' || a[0] == 'l' || a[0] == 'u') b[0] = '3';
    if (a[0] == 'd' || a[0] == 'm' || a[0] == 'v') b[0] = '4';
    if (a[0] == 'e' || a[0] == 'n' || a[0] == 'w') b[0] = '5';
    if (a[0] == 'f' || a[0] == 'o' || a[0] == 'x') b[0] = '6';
    if (a[0] == 'g' || a[0] == 'p' || a[0] == 'y') b[0] = '7';
    if (a[0] == 'h' || a[0] == 'q' || a[0] == 'z') b[0] = '8';
    if (a[0] == 'i' || a[0] == 'r') b[0] = '9';
    if (a[1] == 'a' || a[1] == 'j' || a[1] == 's') b[1] = '1'; 
    if (a[1] == 'b' || a[1] == 'k' || a[1] == 't') b[1] = '2';
    if (a[1] == 'c' || a[1] == 'l' || a[1] == 'u') b[1] = '3';
    if (a[1] == 'd' || a[1] == 'm' || a[1] == 'v') b[1] = '4';
    if (a[1] == 'e' || a[1] == 'n' || a[1] == 'w') b[1] = '5';
    if (a[1] == 'f' || a[1] == 'o' || a[1] == 'x') b[1] = '6';
    if (a[1] == 'g' || a[1] == 'p' || a[1] == 'y') b[1] = '7';
    if (a[1] == 'h' || a[1] == 'q' || a[1] == 'z') b[1] = '8';
    if (a[1] == 'i' || a[1] == 'r') b[1] = '9';
    if (a[2] == 'a' || a[2] == 'j' || a[2] == 's') b[2] = '1';
    if (a[2] == 'b' || a[2] == 'k' || a[2] == 't') b[2] = '2';
    if (a[2] == 'c' || a[2] == 'l' || a[2] == 'u') b[2] = '3';
    if (a[2] == 'd' || a[2] == 'm' || a[2] == 'v') b[2] = '4';
    if (a[2] == 'e' || a[2] == 'n' || a[2] == 'w') b[2] = '5';
    if (a[2] == 'f' || a[2] == 'o' || a[2] == 'x') b[2] = '6';
    if (a[2] == 'g' || a[2] == 'p' || a[2] == 'y') b[2] = '7';
    if (a[2] == 'h' || a[2] == 'q' || a[2] == 'z') b[2] = '8';
    if (a[2] == 'i' || a[2] == 'r') b[2] = '9';
    if (a[3] == 'a' || a[3] == 'j' || a[3] == 's') b[3] = '1';
    if (a[3] == 'b' || a[3] == 'k' || a[3] == 't') b[3] = '2';
    if (a[3] == 'c' || a[3] == 'l' || a[3] == 'u') b[3] = '3';
    if (a[3] == 'd' || a[3] == 'm' || a[3] == 'v') b[3] = '4';
    if (a[3] == 'e' || a[3] == 'n' || a[3] == 'w') b[3] = '5';
    if (a[3] == 'f' || a[3] == 'o' || a[3] == 'x') b[3] = '6';
    if (a[3] == 'g' || a[3] == 'p' || a[3] == 'y') b[3] = '7';
    if (a[3] == 'h' || a[3] == 'q' || a[3] == 'z') b[3] = '8';
    if (a[3] == 'i' || a[3] == 'r') b[3] = '9';
    printf("%s\n",b);
    return 0;
}

Answer 1

假设您的编译器使用ASCII编码，那么您可以使用以下简单算法来获得答案：

1 + (strA[i] - 'a') % 9

你真的不希望用一长串if语句或switch语句来实现它。

如果您有非字母字符，数字字符，大写字符等，您自然会遇到输入验证问题。我认为你可以简单地忽略那些学习练习。

Answer 2

要纠正原始方法，您需要做两件事：

在字符常量周围使用单引号;
使用==检查是否相等;
使用;终止声明。

...所以你的代码段变成：

if (strA[0] == 'a')
    strB[0] = '1';
if (strA[0] == 'b')
    strB[0] = '2';
if (strA[0] == 'c')
    strB[0] = '3';

Answer 3

您可以在gwbasic编辑器中输入以下内容，它将解决您的问题

10 INPUT A$

12 L = LEN(A$)

15 FOR T = 1 TO L

20 M$ = MID$(A$,T,1)

25 GOSUB 70

30 B$ = B$ + V$

35 NEXT T

40 PRINT B$

50 END

55 REM -----------------

70 REM - Subroutine to convert m$ into v$

72 X = ASC(M$) : REM this is the ascii value of m$ (eg. "a" = 97)

74 X = X - 96 : REM so that 97 becomes "1"

80 IF X > 9 THEN X = X - 9 : GOTO 80

90 V$ = STR$(X) : REM just converting to a string type variable

95 RETURN : REM takes you back to line 30 where this value is added to the

96 REM final resulting B$ - when I say added I mean a char added to a string

97 REM such that   "APPL" + "E" = "APPLE"

98 REM ------------------------------------------ DONE

Answer 4

对于ASCII，它有点像：

... make sure strB has enough space ...
for (i = 0; i < ... length of strA ... ; i++) {
    /* you should always check your input is valid */
    if (strA[i] >= 'a' && strB[i] <= 'z')
        strB[i] = (strA[i] - 'a') % 9 + 1;
    else
        strB[i] = ... some default value ...
}

对于EBCDIC：

for (i = 0; i < ... length of strA ... ; i++) {
    /* you should always check your input is valid */
    if (strA[i] >= 'a' && strB[i] <= 'r')
        strB[i] = strA[i] & 0xF;
    else if (strA[i] >= 's' && strB <= 'z')
        strB[i] = (strA[i] & 0xF) - 1;
    else
        strB[i] = ... some default value ...
}

Answer 5

仔细查看ASCII表。您将看到所有字母都使用某个整数值进行编码。首先，如果您只有小写字母，那么从任何字母中删除'a'的字符代码就足以得到您想要的内容

int nr = strA[0] - 'a' + 1;
//now you'd need to convert back to a string; better: strB should be an array of integer

此外，=是赋值运算符;您需要使用==来检查是否相等。

Answer 6

试试这个：）

int strB[MAX_LEN] = {0};
char *strA = malloc (MAX_LEN * sizeof(char));
int i,c = 0,x;

scanf("%s",strA);

for(i = 0 ; i<strlen(strA) ; i++){
    x = strA[i] - 'a' + 1;
    if(x >= 1 && x <= 9)
        strB[c] = x;
    else if(x <= 18){
        strB[c] = x - 10;
    else if(x <= 26){
        strB[c] = x - 19;
    if(x <= 26)
        c++;
}

或者你可以在for循环中使用ninjalj方法，如果你经常检查输入：

for(i=0 ; i<strlen(strA) ; i++){
    strB[i] = (strA[i] - 'a') % 9 + 1;
}

或者这个：

for(i=0 ; i<strlen(strA) ; i++){
    if(strA[i] >= 'a' && strA[i] <= 'z'){
        strB[c] = (strA[i] - 'a') % 9 + 1;
        c++;
    }
}

将字符串数组字符转换为数字

6 个答案: