Question

我一直在研究URLdecode函数，所以我可以解析在HTTP服务器上收到的POST请求，但是我遇到了打嗝。每当我向数组添加整数时，它都不会添加整数的ASCII对应项。有什么我想念的吗？

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void copyAndTerminateString(char *dest, char *src, int len) {
    if (len > 0) strncpy(dest, src, len);
    dest[len] = '\0';
}

char *urldecode(char *encodedString) {
    int i, n = 0, val, encodedStringLen = strlen(encodedString), stringLen = 0;
    for (i = 0; i < encodedStringLen; i++) {
        if (encodedString[i] == '%') {
            stringLen -= 1;
        } else {
            stringLen++;
        }
    }
    char *returnString = malloc(stringLen + 1), *hexValue = malloc(3);
    i = 0;
    printf("length = %d\r\n", stringLen);
    printf("+ is %d\r\n", (int)'+');
    while (encodedString[i] != '\0') {
        if (encodedString[i] == '%') {
            copyAndTerminateString(hexValue, encodedString + i + 1, 2);
            val = strtol(hexValue, NULL, 16);
            if (val != 0) {
                returnString[n] = val;
            }
            printf("ascii \"%s\" = \"%d\"\r\n", hexValue, val);
            i += 2;
        } if (encodedString[i] == '+') {
            returnString[n] = ' ';
        } else {
            returnString[n] = encodedString[i];
        }
        i++;
        n++;
    }
    returnString[stringLen] = '\0';
    return returnString;
}

int main(int argc, char **argv) {
    char *a = "a+%2B+b+%3D%3D+13%25%21";
    char *b = urldecode(a);
    printf("String is: %s\r\n", b);
    free(b);
    return 0;
}

输出：

length = 13
+ is 43
ascii "2B" = "43"
ascii "3D" = "61"
ascii "3D" = "61"
ascii "25" = "37"
ascii "21" = "33"
String is: a B b DD 1351

预期产出：

...
String is: a + b == 13%!

Answer 1

你错过了else行：

        } if (encodedString[i] == '+') {

表示在解码十六进制值后，用最后一个十六进制数字覆盖它......

为什么在此char数组中添加int不会添加相同值的ASCII字符？

1 个答案: