C语言 将 FILE 指针传递给函数

Passing FILE pointer to a function

提问by Sir DrinksCoffeeALot

I'm little bit confused over here, not quite sure about this. What I'm trying to do is to pass the name of a file through terminal/cmdthat will be opened and read from.

我在这里有点困惑，不太确定这一点。我想要做的是通过terminal/传递文件的名称，该文件cmd将被打开和读取。

myfunction(char* fileName, FILE* readFile)
{
    if((readFile = fopen(fileName,"r")) == NULL)
    {
        return FILE_ERROR;
    }
    return FILE_NO_ERROR;
}

int main(int argc, char **argv)
{
FILE* openReadFile;
    if(myfunction(argv[1], openReadFile) != FILE_NO_ERROR)
    {
        printf("\n %s : ERROR opening file. \n", __FUNCTION__);
    }
}

My question is if i pass a pointer openReadFileto myfunction()will a readFilepointer to opened file be saved into openReadFilepointer or do i need to put *readFilewhen opening.

我的问题是，如果我传递一个指针openReadFile，myfunction()将readFile指向打开文件的指针保存到openReadFile指针中，还是*readFile在打开时需要放置。

回答by mksteve

FILE * needs to be a pointer, so in main openReadFile stays as a pointer. myfunction takes **, so we can update the FILE * with the result from fopen *readFile = fopen...updates the pointer.

FILE * 需要是一个指针，所以在主 openReadFile 中仍然是一个指针。myfunction 需要 **，因此我们可以使用 fopen 的结果更新 FILE * *readFile = fopen...更新指针。

int myfunction(char* fileName, FILE** readFile) /* pointer pointer to allow pointer to be changed */
{
    if(( *readFile = fopen(fileName,"r")) == NULL)
    {
        return FILE_ERROR;
    }
    return FILE_NO_ERROR;
}

int main(int argc, char **argv)
{
    FILE* openReadFile; /* This needs to be a pointer. */
    if(myfunction(argv[1], &openReadFile) != FILE_NO_ERROR) /* allow address to be updated */
    {
        printf("\n %s : ERROR opening file. \n", __FUNCTION__);
    }
}