分配使得指针从整数，未投 [英] Assignment makes pointer from integer without cast

问题描述

这是一个Java的背景，我学习C的到来，但我觉得那些模糊的编译器错误消息越来越沮丧。这里是我的code：

Coming from a Java background I'm learning C, but I find those vague compiler error messages increasingly frustrating. Here's my code:

/*
 * PURPOSE
 *      Do case-insensetive string comparison.
 */
#include <stdio.h>
#include <string.h>
#include <ctype.h>

int compareString(char cString1[], char cString2[]);
char strToLower(char cString[]);

int main() {
    // Declarations
    char cString1[50], cString2[50];
    int isEqual;

    // Input
    puts("Enter string 1: ");
    gets(cString1);
    puts("Enter string 2: ");
    gets(cString2);

    // Call
    isEqual = compareString(cString1, cString2);
    if (isEqual == 0)
        printf("Equal!\n");
    else
        printf("Not equal!\n");

    return 0;
}

// WATCH OUT
//      This method *will* modify its input arrays.
int compareString(char cString1[], char cString2[]) {
    // To lowercase
    cString1 = strToLower(cString1);
    cString2 = strToLower(cString2);

    // Do regular strcmp
    return strcmp(cString1, cString2);
}

// WATCH OUT
//      This method *will* modify its input arrays.
char strToLower(char cString[]) {
    // Declarations
    int iTeller;

    for (iTeller = 0; cString[iTeller] != '\0'; iTeller++)
        cString[iTeller] = (char)tolower(cString[iTeller]);

    return cString;
}

这会产生两个警告。

• 赋值时将整数指针，未作投
  
  
  
  • cString1 =用strtolower（cString1）;
  
  
  • cString2 =用strtolower（cString2）;
  
  
  
  
  • 返回CString的;
  
  

  有人能解释这些警告？

  推荐答案

  C字符串不是像Java字符串东西。他们基本上是字符数组。

  C strings are not anything like Java strings. They're essentially arrays of characters.

  您所得到的错误，因为用strtolower返回一个char。一个char是整数C.你是分配给它到一个char []这是一个指针的形式。因此，转换为指针的整数。

  You are getting the error because strToLower returns a char. A char is a form of integer in C. You are assigning it into a char[] which is a pointer. Hence "converting integer to pointer".

  您用strtolower使得其全部到位的变化，没有理由为它返回任何东西，特别是没有一个字符。你应该回归无效，或一个char *。

  Your strToLower makes all its changes in place, there is no reason for it to return anything, especially not a char. You should "return" void, or a char*.

  在调用用strtolower，也没有必要转让，你基本上是刚好路过的存储器地址cString1。

  On the call to strToLower, there is also no need for assignment, you are essentially just passing the memory address for cString1.

  在我的经验，在C字符串是来学习的人从Java / C＃背景回来C.人们可以使用的内存分配相处（因为即便在Java中你经常分配数组）是最难的部分。如果你的最终目标是C ++而不是C，你可能preFER少关注C字符串，请确保您了解的基础知识，只是使用C ++字符串从STL。

  In my experience, Strings in C are the hardest part to learn for anyone coming from Java/C# background back to C. People can get along with memory allocation (since even in Java you often allocate arrays). If your eventual goal is C++ and not C, you may prefer to focus less on C strings, make sure you understand the basics, and just use the C++ string from STL.

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