错误：不兼容的整数到指针转换分配给字符串＆＃x27; （aka＆＃x27; char *＆＃x27;）来自＆＃x27; int＆＃x27; c

发布于 2025-02-03 18:07:43 字数 819 浏览 2 评论 0原文

这是不完整的代码，但我正在尝试制作大写和小写的argv [1]的副本，但要获取错误消息。如果这是一种效率低下的方法，或者通常是完全错误的，我会很感激任何提示，但是在这种特定情况下，有没有办法解决错误？

我可以通过这个问题来确定编码的新手，所以很抱歉，如果这是一个愚蠢的问题，但是错误会在哪里发生呢？我意识到Argv [1]正在转换为整数，但我既不知道在哪里也不知道它如何修复它。

#include <cs50.h>
#include <stdio.h>
#include <string.h>
#include <ctype.h>

int main(int argc, string argv[])
{
    string keyu[26];
    string keyl[26];
    string key = argv[1];

    for (int u = 0; u < strlen(key); u++)
    {
        keyu[u] = toupper(key[u]);
    }

    for (int l = 0; l < strlen(key); l++)
    {
        keyl[l] = tolower(key[l]);
    }

它给出的错误是：

14:17: error: incompatible integer to pointer conversion assigning to 'string' (aka 'char *') from 'int' [-Werror,-Wint-conversion]
keyu[u] = toupper(key[u]);

原文

This is incomplete code but I'm trying to make copies of argv[1] that are uppercase and lowercase, but get an error message. If it's an inefficient way to go about it in general or completely wrong I would appreciate any tips, but is there a way to fix the error in this specific case?

I'm extremely new to coding as you can tell by the question, so I'm sorry if it's a stupid one, but where might the error be occurring? I realize somehow argv[1] is being converted into an integer but I neither know where nor how to really fix it.

#include <cs50.h>
#include <stdio.h>
#include <string.h>
#include <ctype.h>

int main(int argc, string argv[])
{
    string keyu[26];
    string keyl[26];
    string key = argv[1];

    for (int u = 0; u < strlen(key); u++)
    {
        keyu[u] = toupper(key[u]);
    }

    for (int l = 0; l < strlen(key); l++)
    {
        keyl[l] = tolower(key[l]);
    }

The error it gives out is:

14:17: error: incompatible integer to pointer conversion assigning to 'string' (aka 'char *') from 'int' [-Werror,-Wint-conversion]
keyu[u] = toupper(key[u]);

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

始于初秋 2025-02-10 18:07:43

这些声明

string keyu[26];
string keyl[26];

等同于

char * keyu[26];
char * keyl[26];

这些声明，它们是指针的数组。

在这些语句中，

keyu[u] = toupper(key[u]);
keyl[l] = tolower(key[l]);

您正在尝试将整数（例如toupper（key [u]））分配给指针（例如keyu [u]

）看来您实际上要声明字符数组

char keyu[26];
char keyl[26];

要注意的是，您应该将终止零字符'\ 0'在for循环之后附加阵列，以确保数组包含字符串。

These declarations

string keyu[26];
string keyl[26];

are equivalent to

char * keyu[26];
char * keyl[26];

That is they are arrays of pointers.

This in these statements

keyu[u] = toupper(key[u]);
keyl[l] = tolower(key[l]);

you are trying to assign an integer (as for example toupper(key[u])) to a pointer (as for example keyu[u])

It seems you want actually to declare arrays of characters

char keyu[26];
char keyl[26];

Pay attention to that you should append the arrays with the terminating zero character '\0' after the for loops to guarantee that the arrays contain strings.

回复收藏 0 原文

~没有更多了~