警告：赋值丢弃了指针目标类型的限定 [英] warning: assignment discards qualifiers from pointer target type

问题描述

我写了下面code：

void buildArrays(char *pLastLetter[],int length[], int size, const char str[]) {

    int i;
    int strIndex = 0;
    int letterCounter = 0;

    for (i=0; i<size; i++) {

        while ( (str[strIndex] != SEPERATOR) || (str[strIndex] != '\0') ) {
            letterCounter++;
            strIndex++;
        }
        pLastLetter[i] = &str[strIndex-1];
        length[i] = letterCounter;
        letterCounter = 0;
        strIndex++;
    }
}

和我得到的 pLastLetter上述警告[I] =安培; STR [strIndex-1];

pLastLetter是一个数组的指针指向一个char str中[]。

任何人都知道为什么我得到它，以及如何解决它？

Anyone knows why I'm getting it and how to fix it?

推荐答案

嗯，你说的你自己， pLastLetter 是字符数组* 指针，而 STR 是为const char 的数组。在＆放大器; STR [strIndex-1] 前pression的类型为const char * 。您不得指派为const char * 值到的char * 指针。这将违反常量，正确性的规则。其实，你在做什么是C. C编译器的错误报告的传统作为一个单纯的警告，以避免破坏一些老的传统code。

Well, as you said yourself, pLastLetter is an array of char * pointers, while str is an array of const char. The &str[strIndex-1] expression has type const char*. You are not allowed to assign a const char* value to a char * pointer. That would violate the rules of const-correctness. In fact, what you are doing is an error in C. C compilers traditionally report it as a mere "warning" to avoid breaking some old legacy code.

至于如何解决......这取决于你正在尝试做的。要么让 pLastLetter 为const char *的数组或删除常量从 STR 。

As for "how to fix it"... It depends on what you are trying to do. Either make pLastLetter an array of const char* or remove the const from str.

这篇关于警告：赋值丢弃了指针目标类型的限定的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！