Question

问候！

我在qt上有一个简单的程序。有两个指向short类型的指针，用于从文件读取和存储读取值的位。

示例代码：

//(input is a FILE* which is opened and passed to the function)
//(output is also a FILE* which is also opened and passed to the function)

//1. Variables declaration
short* sample_buffer;
int buffer_size=1;
short samples_read;
unsigned long value_x=7;
short* nzb_buffer;
short buffer_position=-1;
int i;

//2.Memory allocation
sample_buffer= malloc(sizeof(short)*buffer_size);
nzb_buffer = malloc(sizeof(short)*value_x);

....

//3. Read from infile, one short at time, process and write it to outfile
do
{
     //3.1. Read from input file
     samples_read = fread(sample_buffer,sizeof(short),buffer_size, input);
     //3.2. Switch position inside nzb_buffer one to the right, 
     //     going back to zero if out of bounds
     buffer_position=(buffer_position+1)%value_x;

     ....

     //3.3. Put least significant bit of the just read short into nzb_buffer
     nzb_buffer[buffer_position]=sample_buffer[0]%2;

     ....

     //3.4. Write the short we just read from infile to the outfile
     for (i=0;i<samples_read;i++)
     {
         fwrite(sample_buffer,sizeof(short),1, output);
     }
} while(samples_read==buffer_size);

我让不可靠的代码片段出来了。如果你需要看别的东西，请告诉我。

问题是，在循环的10或15次操作之后，它会因“分段故障”信号而崩溃。它在fwrite（）函数上崩溃。

我调试了，我在sample_buffer上使用watch。由于某种原因，在一个确切的步骤，操作nzb_buffer [buffer_position] = sample_buffer [0]％2使sample_buffer变为0x0（我相信，它变成空指针）。

这不能在nzb_buffer上溢出，因为该操作的buffer_position是3（在malloc中为特定数组分配的7个）。并且由于每个循环进行一次写操作并移位进位，因此写入nzb_buffer [3]的操作在循环之前就已经发生，并且没有使指针无效。

我完全不知道这里可能发生什么。任何人都有任何想法发生了什么或如何调试它？

提前致谢！

PS：添加了评论“代码的作用”

Answer 1

循环的退出条件似乎是错误的。我愿意：

samples_read = fread(sample_buffer,sizeof(short),buffer_size, input);
while(samples_read==buffer_size){

    [...]        

    samples_read = fread(sample_buffer,sizeof(short),buffer_size, input);
}

由于没有明显的原因，指针变得毫无意义

1 个答案: