我的gcc編譯器給我一個錯誤，顯示為（函數）的沖突類型

Question

我是C語言的新手，我制作了此程序，但出現錯誤“（函數）的類型沖突”“（函數）的先前聲明在這里”。

我在系統上使用命令提示符使用Dev c ++的gcc編譯器對此進行了編譯。 有人可以幫我了解我的錯嗎？

#include<stdio.h>
#include<math.h>
main()
{
    int a,b,c;
    float area;
    float ar(int a,int b,int c);
    printf("Enter the lenghts of the three sides of a triangle");
    scanf("%d%d%d",&a,&b,&c);
    area=ar(a,b,c);
    printf("The are a of the triangle is=%.2f",area);
}
ar(a,b,c)
{
    float area,s;
    s=(a+b+c)/3;
    area=sqrt((s*(s-a)*(s-b)*(s-c)));
    return area;
}

Answer 1

您使用返回類型float聲明了函數“ ar”，但返回了float ar（int a，int b，int c）; 但您沒有定義返回值。這在這里引起了問題。 嘗試這個：

    #include<stdio.h>
    #include<math.h>
    float ar(int a,int b,int c);

    void main()
    {
        int a,b,c;
        float area;
        printf("Enter the lenghts of the three sides of a triangle");
        scanf("%d%d%d",&a,&b,&c);
        area=ar(a,b,c);
        printf("The are a of the triangle is=%.2f",area);
    }
    float ar(int a,int b,int c)
    {
        float area,s;
        s=(a+b+c)/3;
        area=sqrt((s*(s-a)*(s-b)*(s-c)));
       return area;
   }

使用命令：gcc -std = c99 -o stack_16_4_16 stack_16_4_16.c -lm編譯

Answer 2

問題在於您在原型中指定了返回類型，但未在實際函數定義中指定返回類型。 當使用特定定義編寫原型時，編寫實際函數時必須遵循相同的定義。

我還將整數加法轉換為浮點數，以便可以正確計算面積。 當您使用整數並將其除以C中的浮點數時，無論您將其分配給什么，都將得到一個整數。 演員表將改變這種行為。

考慮一下：

#include<stdio.h>
#include<math.h>
int main()
{
    int a,b,c;
    float area;
    float ar(int a,int b,int c);
    printf("Enter the lenghts of the three sides of a triangle");
    scanf("%d%d%d",&a,&b,&c);
    area=ar(a,b,c);
    printf("The are a of the triangle is=%.2f",area);
}
float ar(int a,int b,int c)
{
    float area,s;
    s=(float)(a+b+c)/3;
    area=sqrt((s*(s-a)*(s-b)*(s-c)));
    return area;
}

這將干凈地編譯。

Answer 3

我試圖在這里包括一些解釋：

#include<stdio.h>
#include<math.h>

float ar(int a,int b,int c); 
/*  This is a function declaration and just like variables, functions are
 *  also limited by scope. If you declare the function inside the main,
 *  then the function cannot be called outside the main. That is the purpose
 *  it is declared here.
 */

int main()
{
    int a,b,c;
    float area;

    printf("Enter the lengths of the three sides of a triangle : ");
    scanf("%d%d%d",&a,&b,&c);
    area=ar(a,b,c);
    printf("The are a of the triangle is = %.2f",area);

    return 0;
}

    float ar(int a,int b,int c)  
    /* Note that I have added the types of arguments
     * In older versions of C it is not uncommon to see functions like 
     * float ar(a,b)
     * int a,b;
     * {Some Stuff Here}
     * The above function style was problematic - it couldn't deal with mismatched arguments.
     * So it is a good practice to specify the type of the parameters.
     * In fact this was ANSI C standard's solution to the problems of mismatched arguments.
     */
    {
    float area,s;
    s=(a+b+c)/(float)2;  
    /* Either the numerator or the denominator  should be float for the output to be float
     * So I casted the denominator to a float value. Also, if you're using Heron's formula
     * the denominator should be 2 , not 3.
     */

    area=sqrt(s*(s-a)*(s-b)*(s-c));

    return area;
    }

Answer 4

您應該在函數ar中添加返回類型

#include<stdio.h>
#include<math.h>
main()
{
    int a,b,c;
    float area;
    float ar(int a,int b,int c);
    printf("Enter the lenghts of the three sides of a triangle");
    scanf("%d%d%d",&a,&b,&c);
    area=ar(a,b,c);
    printf("The are a of the triangle is=%.2f",area);
}
float ar(float a, float b, float c)
{
    float area,s;
    s=(a+b+c)/3;
    area=sqrt((s*(s-a)*(s-b)*(s-c)));
    return area;
}