为什么调用此函数会产生错误“<function> 不是函数或函数指针”？

Question

似乎函数 qwertyInches() 应该可以工作但是当我在 main() 中调用它时它给了我

[Error] called object 'qwertyInches' is not a function or function pointer.

如有任何帮助，我们将不胜感激。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// Global constant
const int MAX_LENGTH = 81;   // global constant for max input line length


void qwertyInches (char row[], double *inches, int x, double y) {
    int d;
    for (d = 0; d < strlen(row); d++) {
        if (x == row[d]) {
            *inches = *inches + y;
        }
    }
}


int main() {
    int count[256] = { 0 };
    int  letterCounter = 0;
    int qwertyCounter = 0;
    int homeRowCounter = 0;
    int dvorakCounter = 0;
    char qwertyHomeRow[23] = {"asdfghjkl;ASDFGHJKL:\"'" };
    char dvorakHomeRow[22] = {"aoeuidhtns-_AOEUIDHTNS"};
    double percentOfCharQwerty = 0.0;
    double percentOfCharDvorak = 0.0;
    char qwertyHomeRowInches[4] = {"ghGH"};
    char qwertyRowInches[46] = {"qweruiopQWERUIOP{[}]\ZzXx|CcVvNnMm<,>./?"};
    char qwertyNumberInches[25]= {"`~1!2@3#4%7&8*9(0)-_=+)"};
    char qwertyTAndYInches[4] = {"TtYy"};
    char num6Inches[2] = {"6^"};
    char dvorakHomeRowInches[4]= {"iIDd"};
    char dvorakRowInches[41] = {"\"<',.>PpGgCcRrLl?/:+=|:;QqJjKkBb\MmWwVvZz"};
    char dvorakYandFInches[4] = {"YyFf"};
    char dvorakNumberInches [25] = {"~`1!2@3#4%7&8*9()0{[]}"};
    double dvorakInches = 0.0;
    double qwertyInches = 0.0;


    /* loop counters */
    int k;
    int l;
    int d;
    /* file handle --- in this case I am parsing this source code */
    FILE *fp = fopen("newsample.txt", "r");

    /* a holder for each character (stored as int) */
    int c;

    /* for as long as we can get characters... */
    while((c=fgetc(fp))) {

        /* break if end of file */
        if(c == EOF) {
            break;
        }
        else if (c == 32 || c == 10 || c == 9) {
            count[c]+=1;
        }
        /* otherwise add one to the count of that particular character */
        else {
            count[c]+=1;
            letterCounter++;
            for (l = 0; l < strlen(dvorakHomeRow); l++) {
                if (c == qwertyHomeRow[l]) {
                    qwertyCounter++;
                }
                if (c == dvorakHomeRow[l]) {
                    dvorakCounter++;
                }
            }
            qwertyInches(strlen(qwertyHomeRowInches) , &qwertyInches, c, .75 );

        }

    }


percentOfCharQwerty = (double) qwertyCounter / (double) letterCounter * 100;
percentOfCharDvorak = (double) dvorakCounter / (double) letterCounter * 100;

printf("Amount of Letters: %d\n", letterCounter);
printf("qwerty counter: %d\n", qwertyCounter);
printf("Dvorak counter: %d\n", dvorakCounter);
printf("Percent of qwerty letters %.2lf\n", percentOfCharQwerty);
printf("Percent of Dvorak letters %.2lf\n", percentOfCharDvorak);
printf("qwertyInches: %.2lf\n", qwertyInches);
printf("dvorakInches: %.2lf\n", dvorakInches);
/* close the file */
fclose(fp);
return;
}

Answer 1

您将 qwertyInches 定义为变量和函数。

void qwertyInches (char row[], double *inches, int x, double y) {

和

   double qwertyInches = 0.0;

只需更改以上其中一项的名称即可。我通常给我的函数起一个 "action" 的名字，给我的变量起一个 "thing" 的名字。

Answer 2

when i call it in main it gives me this... [Error] called object 'qwertyInches' is not a function or function pointer.

当然编译器是对的。 在 main() 中，这个声明隐藏了函数声明：

double qwertyInches = 0.0;

因此，在main()中，qwertyInches指的是double类型的局部变量，而不是函数。

Answer 3

main() 中有一个 qwertyInches 局部变量，它隐藏了 qwertyInches() 函数。

引用 C11，章节 §6.2.1，标识符的范围（强调我的）

[....] If an identifier designates two different entities in the same name space, the scopes might overlap. If so, the scope of one entity (the inner scope) will end strictly before the scope of the other entity (the outer scope). Within the inner scope, the identifier designates the entity declared in the inner scope; the entity declared in the outer scope is hidden (and not visible) within the inner scope.

解决方法：改其中一个名字

就是说，qwertyInches() 函数的第一个参数应该是 char *，但是您传递的是 size_t（strlen() 的输出），这很简单错误的。也改变它。

为什么调用此函数会产生错误“<function> 不是函数或函数指针”？

Why calling this function produces error "<function> is not a function or function pointer"?

c

function

scope

function-call