I am currently in process of making an image-processing program and it is going pretty well but suddenly some fopen() errors appeared. In my previous version (no dynamically allocated arrays, no structs) it was okay, char name[50] was used in fopen(name, "a") and it worked good but now I cannot make it work at all. Whatever I throw in fopen(), it does not compile. Here is full error description:
我目前正在制作一个图像处理程序，它运行得很好，但是突然出现了一些fopen()错误。在我以前的版本中(没有动态分配的数组，没有struct)，它是可以的，char[50]在fopen(名称，“a”)中使用，它工作得很好，但是现在我不能让它工作了。无论我在fopen()中抛出什么，它都不会编译。这里是完整的错误描述:
obsluga.c:30:8: error: incompatible types when assigning to type ‘FILE’ from type ‘struct FILE *’ plik2=fopen(name, "a");
obsluga。c:30:8:错误:不兼容的类型，当从类型' struct文件* ' plik2=fopen(名称，"a")中指定类型' FILE '时，不兼容的类型;
code fragment (I could post more but it should be irrelevant)
代码片段(我可以发布更多，但它应该是无关的)
char name[50];
plik2=fopen(name, "a");
What I really want is to pass to fopen() file name contained in char *output which is in the appropriate structure but sadly even the easiest way doesn't work right now.
我真正想要的是将char *输出中包含的fopen()文件名传递到适当的结构中，但遗憾的是，即使是最简单的方法也不能立即生效。
1 个解决方案
#1
2
To get the error, you must have written something like:
为了得到错误，你一定写了这样的东西:
FILE plik2;
instead of the correct:
而不是正确的:
FILE *plik2;
Treat FILE * as an opaque type. You always write FILE * (almost always; occasionally, you might use FILE **) and never write FILE.
将文件*视为不透明类型。你总是写文件*(几乎总是;偶尔，您可能会使用文件**)，而从不写文件。
#1
2
To get the error, you must have written something like:
为了得到错误，你一定写了这样的东西:
FILE plik2;
instead of the correct:
而不是正确的:
FILE *plik2;
Treat FILE * as an opaque type. You always write FILE * (almost always; occasionally, you might use FILE **) and never write FILE.
将文件*视为不透明类型。你总是写文件*(几乎总是;偶尔，您可能会使用文件**)，而从不写文件。