递归函数中的内存泄漏

Question

这是我在C中实现Cooley-Tukey算法的片段.是的,它是大学作业.但无论如何......算法工作正常,但我必须释放ar1和ar2来摆脱巨大的输入数据上的巨大内存泄漏,但每次我尝试,我得到无效的读取.理论上,ar1和ar2应仅由函数的当前实例使用,并且它们应该是唯一的,因为每个实例malloc都有自己的输出.

complex_exp * dft(complex_exp * from, int N, int s, int inverse) {

if(N == 1)
    return from;

int i;
complex_exp * transformed = (complex_exp *) malloc(N * sizeof(complex_exp));
complex_exp * ar1 = dft(from, N / 2, 2*s, inverse); //LINE 83
complex_exp * ar2 = dft(from + s, N / 2, 2*s, inverse); // LINE 84

for(i = 0; i < N/2; i++) {

    transformed[i] = ar1[i]; //LINE 88
}

for(i = N/2; i < N; i++) {
    transformed[i] = ar2[i - N/2];
}

//Do stuff with the transformed array - NO reference to ar1 or ar2.

free(ar1); //LINE 113
return transformed;
}

Valgrind说:

==69597== Invalid read of size 8
==69597==    at 0x100000EE6: dft (progtest05.c:88)
==69597==    by 0x100000EA2: dft (progtest05.c:84)
==69597==    by 0x100000E67: dft (progtest05.c:83)
==69597==    by 0x100000E67: dft (progtest05.c:83)
==69597==    by 0x100001A0E: main (progtest05.c:233)
==69597==  Address 0x100007250 is 64 bytes inside a block of size 256 free'd
==69597==    at 0xDCB8: free (vg_replace_malloc.c:450)
==69597==    by 0x1000011E5: dft (progtest05.c:113)
==69597==    by 0x100000E67: dft (progtest05.c:83)
==69597==    by 0x100000E67: dft (progtest05.c:83)
==69597==    by 0x100000E67: dft (progtest05.c:83)
==69597==    by 0x100001A0E: main (progtest05.c:233)

因此,似乎第83行的dft调用释放了内存,然后下一行的dft调用尝试访问.知道实际发生了什么以及如何摆脱泄漏？

Answer 1

你说"每个实例mallocs都有自己的输出",但是在这个声明中并不是这样:

if(N == 1)
    return from;

也许当N == 1时你应该返回一个副本from(即malloc新内存,将内容复制到新内存中,然后返回副本).

我怎么能摆脱泄漏？

我希望你必须在返回转换之前释放ar1和ar2.