冲突类型错误 [英] Conflicting Types Error

问题描述

当我运行以下代码时，出现错误

when i run the following code, i get the error

 problem1.c:136:6: error: conflicting types for 'strcspn' 
   int strcspn( char * str, char * reject ) 

我不知道为什么会出现冲突类型错误.这是代码:

I'm not sure why I am getting the conflicting types error. here is the code:

int strcspn( char * str, char * reject ) 
{

    int counter = 0;

    for (int i = 0; i < strlen(str); i ++)
    {       for (int j = 0; j < strlen(reject); j++)
                if ( *(str + i) == *(reject + j) )
                return counter;
        counter++;
    }

return counter;

}


void main ()
{


char * str1 = (char *)malloc(sizeof(char)*100);
char * str2 = (char *)malloc(sizeof(char)*100);
sprintf(str1, "abc123");
sprintf(str2, "d2");
printf("%d\n", strcspn(str1, str2)); 

}

推荐答案

正如 lowtech 在他的回答中已经说过的，你应该避免在你的 C 程序中重新定义已经被采用的函数的名称.

As lowtech already said in his answer, you should avoid to redefine names of the functions in your C programs that are already taken.

无论如何，您的程序存在一些您应该知道的问题.

Any way there are some issue with your program which you should know.

1) strlen 的返回类型是 size_t 而不是 int.
2)主要至少应该是 int main(void){}
3)没有必要投malloc，它的返回类型是void*
4)最重要的是，你应该总是释放你 malloc 的东西.

1) strlen's return type is size_t and not int.
2) main should be at least int main(void){}
3) there is no need to cast malloc, its return type is void*
4) and most important one, you should always free what you malloc.

看看这里:

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

int strcspn_ss( char * str, char * reject ){
    int counter = 0;
    size_t i,j;

    for (i = 0; i < strlen(str); i ++){
        for (j = 0; j < strlen(reject); j++)
                if ( *(str + i) == *(reject + j) )
                return counter;
        counter++;
    }

    return counter;
}


int main (void){

    char * str1 = malloc(sizeof(char)*100);
    char * str2 = malloc(sizeof(char)*100);
    sprintf(str1, "abc123");
    sprintf(str2, "d2");
    printf("%d\n", strcspn_ss(str1, str2));

    free(str1);
    free(str2);
    return 0;
}

就像 cad 在他的评论中所说的，有一个重要的事情你应该知道，如果你在函数内部声明一个变量或用作参数不会影响函数 strcspn，请参阅以下内容:

like cad said in his comment, there is an Important thing which you should know, that if you declare a variable inside a Function or used as parameter doesn't affect the function strcspn, please see the following:

#include<stdio.h>

void foo(int strcspn){
    printf("strcspn = %d\n",strcspn);
}


int main (void){
    int strcspn = 10;
    foo(strcspn);
    return 0;
}

这是合法的.

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