如何修复不兼容的指针类型传递＆＃39; char [16]＆＃39;参数类型＆＃39; FILE *＆＃39; （又名＆＃39; struct __sFILE *＆＃39;）[ - 兼容指针类型]

Question

如何修复？

  

警告：

     

不兼容的指针
           类型传递＆＃39; char [16]＆＃39;参数类型＆＃39; FILE *＆＃39;
           （又名＆＃39; struct __sFILE *＆＃39;）[ - 兼容指针类型]
         的fread（缓冲器，1512，数据）;
                         ^ ~~~
  /Applications/Xcode.app/Contents/Developer/Platforms/MacOSX.platform/Developer/
  SDKs / MacOSX10.13.sdk / usr / include / stdio.h:247:90：注意：
       将参数传递给参数＆＃39; __ stream＆＃39;这里
   ...__ ptr，size_t __size，size_t __nitems，FILE * __restrict __stream）;
                                                                ^
  生成了1个警告。

这是我的代码：

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>

int main()
{

    char readDATA[64];
    char buffer[8];
    char data[16];

    FILE *fl = fopen(data,"r");
    fgets(readDATA,64,fl);
    fread(buffer,1,512,data);
    printf("%s",readDATA);

    return 0;
}

我尝试打开一条路径或随机播放＆＃34;请帮忙。

Answer 1

根据fread()

的手册页

  

函数fread（）读取数据的nmemb元素，每个字节大小          从流指向的流，将它们存储在位置          由ptr。

给出

size_t fread(void *ptr, size_t size, size_t nmemb, FILE *stream);

fread()的最后一个参数是FILE *stream，表示您想要读取内容并存储到buffer的文件。

替换

 fread(buffer,1,512,data); /* last argument is wrong */

与

fread(buffer,1,512,f1);

同时检查fopen()

的返回值

FILE *f1 = fopen("data","r"); /* it will try to open a file called "data" in current working directory, you can take input from user also */
if(f1 == NULL) {
  fprintf(stderr, "file not present\n");
  return 0;
}

你的意图可能是这样的

char data[16];/* it doesn't contain anything, so take the input from user or assign directly */
printf("enter the file name\n");
scanf("%s",data);
FILE *f1 = fopen(data,"r"); 
    if(f1 == NULL) {
      fprintf(stderr, "file not present\n");
      return 0;
    }