如何将字符串转为整数？40位十六进制字符串转Int40结构体求助

Hey there, let's work through this problem step by step. Your current code has a few critical issues that are keeping it from working correctly, so we'll fix those first, then implement the proper way to parse that 40-digit hex string into your Int40 structure.

First, Let's Spot the Problems in Your Existing Code #

Your current approach has three key mistakes:

1. Incorrect memory allocation: You're allocating space for MAX40 instances of Int40, but you only need a single Int40 struct. The struct's digits pointer is what needs to point to a dynamically allocated array of MAX40 integers.
2. Wrong character-to-integer conversion: Casting str[i] directly to int gives you the ASCII value of the character (e.g., '0' becomes 48, not 0). You need to convert each hex character to its actual numeric value (0-15).
3. No reverse storage: The struct requires digits to be stored in reverse order, but you're copying characters in the same order as the string.

Correct Implementation of parseString #

First, make sure you define MAX40 as 40 (since we're dealing with a 40-digit hex number):

#define MAX40 40

Here's the fixed parseString function, with proper memory management, character conversion, and reverse storage:

#include <stdlib.h>
#include <ctype.h> // For tolower()

typedef struct Int40 {
    int *digits;
} Int40;

Int40 *parseString(char *str) {
    // Allocate memory for the Int40 struct itself
    Int40 *value = malloc(sizeof(Int40));
    if (!value) {
        return NULL; // Handle memory allocation failure
    }

    // Allocate memory for the digits array (40 integers)
    value->digits = malloc(sizeof(int) * MAX40);
    if (!value->digits) {
        free(value); // Clean up the struct if digits allocation fails
        return NULL;
    }

    // Iterate through the string, storing digits in reverse order
    for (int i = 0; i < MAX40; i++) {
        // Get the character from the string in reverse order
        char c = str[MAX40 - 1 - i];
        int digit;

        // Convert hex character to its numeric value
        if (isdigit(c)) {
            digit = c - '0'; // '0' to '9' become 0-9
        } else {
            c = tolower(c); // Handle uppercase/lowercase letters uniformly
            if (c >= 'a' && c <= 'f') {
                digit = 10 + (c - 'a'); // 'a'-'f' become 10-15
            } else {
                // Invalid hex character: clean up and return error
                free(value->digits);
                free(value);
                return NULL;
            }
        }

        value->digits[i] = digit;
    }

    return value;
}

How to Use This in main #

Here's an example main function to test the parser, including memory cleanup to avoid leaks:

#include <stdio.h>

int main() {
    // Example 40-digit hex string (replace with your actual input)
    char hexStr[MAX40 + 1] = "0123456789abcdef0123456789abcdef01234567";
    Int40 *p = parseString(hexStr);

    if (!p) {
        printf("Failed to parse the hex string (invalid characters or memory issue)\n");
        return 1;
    }

    // Verify the result: print the digits in reverse storage order
    printf("Stored digits (reverse of input string):\n");
    for (int i = 0; i < MAX40; i++) {
        printf("%d ", p->digits[i]);
    }
    printf("\n");

    // Always free allocated memory to avoid leaks
    free(p->digits);
    free(p);

    return 0;
}

Key Details on Hex String to Integer Conversion #

Let's break down the character-to-integer logic since that's a core part of your question:

• Numeric characters ('0'-'9'): Subtract the ASCII value of '0' from the character. Since '0' is ASCII 48, '5' (ASCII 53) minus '0' gives 5, which is the correct numeric value.
• Alphabetic characters ('a'-'f' or 'A'-'F'): Convert to lowercase first, then subtract 'a' (ASCII 97) and add 10. For example, 'F' becomes 'f', 'f'-'a' is 5, plus 10 gives 15, which is the hex value of F.
• Error handling: If we encounter a character that's not a valid hex digit, we clean up any allocated memory and return NULL to signal failure.

内容的提问来源于stack exchange，提问作者pteran