C - 传递 ‘fprintf’ 的参数 1 从整数生成指针而不进行强制转换

Question

非常简单的脚本抛出这个错误:

passing argument 1 of ‘fprintf’ makes pointer from integer without a cast

这是为什么呢？代码是:

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

int main(int argc, char *argv[])
{
FILE *FP;

if((FP = fopen("file.txt", "r+")) == NULL) {
    printf("File won't open\n");
    return(1);
}

foo(FP);

return(0);
}

int foo(FP) {

    char name[31];
    printf( "Please enter a name (within 30 char) : \n");
    gets(name);
    fprintf(FP, "%s\n", name);

    return(0);
}

Answer 1

改变

int foo(FP) {

到

int foo(FILE * FP) { 

您必须将 foo 函数的原型(prototype)放在代码的顶部。