在C语言中“有太多的参数无法发挥作用”

I define a function

我定义一个函数

int find(char *t, int len){
}

then i call it with

然后我用

value = "hello world";
rt = find(value, strlen(value));

it does not work, and show "error: too many arguments to function ‘find’"

它不起作用，并且显示了“错误:太多的参数不能用于‘查找’”

4 个解决方案

#1

int find(char *t, int len){
}

might give a warning that function should return a value.

可能会提示函数应该返回一个值。

and if you add:

如果你添加:

char* value = "hello world";
int rt = find(value, strlen(value));

It should work fine if the code is in a single file (as already pointed by Michael in comments) else you will have to specify the prototype of find function before calling it from a separate file.

如果代码在一个文件中(如Michael在注释中指出的)，那么它应该可以正常工作，否则您必须在从单独的文件调用之前指定find函数的原型。

#2

There is a syntax error in your call, no ; after value = "hello world"

您的调用有语法错误，没有;after value = "hello world"
Did you #include <string.h>?

你# include < string.h >吗?

#3

This error might occur when there is difference in arguments between function declaration and function definition.

当函数声明和函数定义之间的参数不同时，可能会出现此错误。

#4

There are two errors that I found in the above code .

我在上面的代码中发现了两个错误。

You have to mention the 'return' keyword at the end of the function definition .

您必须在函数定义的末尾提到“return”关键字。

You have to declare the character pointer (char * value) while initialize the 'value' .

您必须在初始化“值”时声明字符指针(char * value)。

#1

int find(char *t, int len){
}