未初始化的值是由堆栈分配 Valgrind c++ 创建的

Question

所以，当运行命令：--track-origins=yes 在 Valgrind

上时，我得到这个错误

==57680== Conditional jump or move depends on uninitialised value(s)
==57680==    at 0x100584A6B: __dtoa (in /usr/lib/system/libsystem_c.dylib)
==57680==    by 0x1005AB2DD: __vfprintf (in /usr/lib/system/libsystem_c.dylib)
==57680==    by 0x1005D1AD2: __v2printf (in /usr/lib/system/libsystem_c.dylib)
==57680==    by 0x1005AA2B1: vfprintf_l (in /usr/lib/system/libsystem_c.dylib)
==57680==    by 0x1005A841D: printf (in /usr/lib/system/libsystem_c.dylib)
==57680==    by 0x100002466: main (test.cpp:500)
==57680==  Uninitialised value was created by a stack allocation
==57680==    at 0x100002025: main (test.cpp:336)

第 500 行是：

  printf("Iteration = %d    t = %.10f   p_iter = %d\n", iter, time[iter+1], p_iter);

第 336 行是：

    double time[iter_max + 1];

我不知道这是什么意思？这次是如何造成内存泄漏问题的？

主要代码示例

#include<math.h>
#include<stdio.h>
#include "Functions.h"
#include "fftw3.h"
#include <cstring>

int main(){         
    
    // Setting parameters 
    double frequency = 200.0;    
    double dt = 0.1;        
    double tend = 5000.; 
    double err_max = 1e-7;          

    int P_iter_max = 500;      

    int iter_max = tend / dt;
    int saveNumber = 1;

    //Initialize Parameters 

    double *ksqu; 
    ksqu = (double*) fftw_malloc(nx*nyk*sizeof(double));
    memset(ksqu, 42, nx*nyk* sizeof(double)); //initialize for now

    fftw_complex *Pek;
    Pek = (fftw_complex*) fftw_malloc(nx*nyk* sizeof(fftw_complex)); 
    memset(Pek, 42, nx*nyk* sizeof(fftw_complex)); 

    fftw_complex *Pik;
    Pik = (fftw_complex*) fftw_malloc(nx*nyk* sizeof(fftw_complex)); 
    memset(Pik, 42, nx*nyk* sizeof(fftw_complex)); 

    fftw_complex *nuink;
    nuink = (fftw_complex*) fftw_malloc(nx*nyk* sizeof(fftw_complex)); 
    memset(nuink, 42, nx*nyk* sizeof(fftw_complex)); 

    //similarly initialize everything else

    //Save I.Cs 

    int p_iter = Potk(invnk, dndxk, dndyk, Pk, PSk, kx, ky, ninvksq, err_max, P_iter_max);

    double time[iter_max + 1]; 
    time[0] = 0;

    for (int iter = 0; iter < iter_max; iter++){

        calcPSk(dndxk, dndyk, Pik, Pek, nuink, nuiek, nuenk, nueik, isigPk, Oci, Oce, u, B, ksqu, potSourcek);
 
    }

    printf("Iteration = %d    t = %.10f   p_iter = %d\n", iter, time[iter+1], p_iter);
}

Answer 1

这不是泄密。您正在读取未初始化的内存。正如@PaulMcKenzie 在评论中所说，time 不是标准的 C++，而是 GCC extension that comes from C99.

您可以使用 memset 来确保 time 已初始化。或者，您可以使用 std::vector，例如

std::vector<double> time(iter_max + 1, 0.0);

这将创建一个大小为 iter_max + 1 的向量，全部初始化为 0.0。

但是，就目前而言，除了在 printf 中，您的代码没有使用 time 变量，因此您可以简单地删除它。

未初始化的值是由堆栈分配 Valgrind c++ 创建的

Uninitialised value was created by a stack allocation Valgrind c++

c++

valgrind