C语言 “错误：在数字常量之前预期为 ';'、'、' 或 ')'”出现在我的代码中
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"error: expected ';', ',' or ')' before numeric constant" is coming up to my code
提问by madLad
The program is about function making an average. I get an error:
该程序是关于函数求平均值。我收到一个错误：
error: expected ';', ',' or ') before numeric constant
错误：在数字常量之前应有 ';'、',' 或 ')
within the avg_array() function whenever I build it. Help would be appreciated, thanks!
每当我构建它时，在 avg_array() 函数中。帮助将不胜感激，谢谢！
#include <stdio.h>
#define SIZE 5
// Prototypes
int avg_array (int*, int);
main()
{
int values[SIZE];
int avg;
int i;
printf("Enter 5 numbers please. \n");
for(i=0; i<SIZE; i++)
{
scanf("%d", &values[i]);
}
avg = avg_array(values, SIZE);
printf("\n The avg of the array is %d \n", avg);
getchar();
getchar();
} // end main()
/* Implement avg_array() WHERE THE ERROR PERTAINS */
avg_array(int my_array[], int SIZE)
{
int sum;
int i;
int fxn_average;
for(i=0; i<SIZE; i++)
{
sum = sum + my_array[i];
}
fxn_average = (sum/SIZE);
return (fxn_average);
}
回答by e0k
You are using the identifier SIZEas an argument. This is also a macro that gets converted to 5by the preprocessor. After the preprocessor applies the macros, it would look like
您正在使用标识符SIZE作为参数。这也是一个被5预处理器转换成的宏。在预处理器应用宏之后，它看起来像
avg_array (int my_array[], int 5)
Since 5is a numeric constant instead of an identifier, it generates an error. Change the variable name.
由于5是数字常量而不是标识符，因此会产生错误。更改变量名称。
It looks like you also have a function signature missing a return type, which should match the prototype declared above. Try this instead:
看起来您还有一个缺少返回类型的函数签名，它应该与上面声明的原型相匹配。试试这个：
int avg_array (int *my_array, int size)
{
int sum = 0;
int i;
for(i=0; i<size; i++)
{
sum = sum + my_array[i];
}
return sum/size;
}
The variable sumshould be initialized to 0. The local variable fxn_averageis not needed because you can use just return sum/size;at the end instead.
该变量sum应初始化为 0。fxn_average不需要局部变量，因为您可以只return sum/size;在最后使用。
I changed the type of the first argument from int[](array of int) to int *(pointer to int) so the function definitionmatches the prototype given in the question. The function was declaredas
我将第一个参数的类型从int[](array of int) 更改为int *(pointer to int)，因此函数定义与问题中给出的原型相匹配。该函数被声明为
int avg_array (int*, int);
These arguments have no identifiers; only their types are specified. This is valid C, but many style guides prescribe against it since naming arguments helps the reader understand meaning or intent. If you are writing a programming interface, for example, all the programmer will likely see is the function prototypes in a header file. It must be clear what the arguments are to write a correct function call. Anyway, adding in identifiers looks like:
这些参数没有标识符；仅指定了它们的类型。这是有效的 C，但许多风格指南都反对它，因为命名参数有助于读者理解含义或意图。例如，如果您正在编写编程接口，那么程序员可能看到的只是头文件中的函数原型。必须清楚参数是什么才能编写正确的函数调用。无论如何，添加标识符看起来像：
int avg_array (int *my_array, int size);
which is the same as in the definitionI used above.
这与我上面使用的定义相同。
