我需要实现延迟输出。我试过这样
def draw(window):
window.scene.addLine(10, 10, 300, 300, pen=window.pen)
sleep(2)
window.scene.addLine(59, 60, 200, 100, pen=window.pen)
sleep(2)
window.scene.addLine(100, 10, 500, 500, pen=window.pen)
sleep(2)
window.scene.addLine(10, 10, 440, 400, pen=window.pen)
但它没有按预期工作。程序只是停止 N 秒，然后立即显示所有行。我该如何解决？
为什么执行我的程序时会忽略多项式 B 读取？（第 3 行）如何解决这个问题？
case (SUM):
puts("Enter polynom B: ");
fgets(str_2, 255, stdin);
rc = get_polynom(str_2, &lst_2);
//print_list(lst_2);
puts("");
polynom_addiction(&lst, &lst_2, &res_list);
print_list(res_list);
break;
case (DVD):
break;
}
主要开始：
int main()
{
int rc = OK;
char choice[255];
char str[255];
char str_2[255];
node_t *lst = NULL;
node_t *lst_2 = NULL;
node_t *res_list = NULL;
double a;
puts("Enter polynom: ");
fgets(str, 255, stdin);
rc = get_polynom(str, &lst);
print_list(lst);
puts("CHOSE THE ACTION: val; ddx; sum; dvd");
fgets(choice, 4, stdin);
你好！编写了一个从二维数组中删除字符串的程序。一切似乎都正常工作，但 valgrind 显示内存丢失和错误。请告诉我我做错了什么。
#include "cdio.h"
#include "func.h"
int del(long int **matrix, size_t *n, int pos)
{
free(matrix[pos]);
for (size_t i = pos; i < *n-1; i++)
matrix[i] = matrix[i+1];
*n -= 1;
long int **temp = realloc(matrix, *n * sizeof(long int *));
if (temp)
{
matrix = temp;
temp = NULL;
}
return SUCCESS;
}
int main()
{
long int **matrix = NULL;
size_t n = 5;
size_t m = 4;
matrix = matrix_allocate(n, m);
for (size_t i = 0; i < n; i++)
for (size_t j = 0; j < m; j++)
matrix[i][j] = rand() % 100;
puts("OLD MATRIX:");
print_matrix(matrix, n, m);
int pos = 0;
puts("enter position:");
scanf("%d", &pos);
del(matrix, &n, pos);
puts("NEW MATRIX:");
print_matrix(matrix, n, m);
for (int i = 0; i < n; i++)
free(matrix[i]);
free(matrix);
return 0;
}
OLD MATRIX:
83 86 77 15
93 35 86 92
49 21 62 27
90 59 63 26
40 26 72 36
enter position:
1
NEW MATRIX:
==3691== Invalid read of size 8
==3691== at 0x108E30: print_matrix (cdio.c:60)
==3691== by 0x108B86: main (del.c:47)
==3691== Address 0x522d040 is 0 bytes inside a block of size 40 free'd
==3691== at 0x4C31D2F: realloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==3691== by 0x108A1C: del (del.c:12)
==3691== by 0x108B63: main (del.c:43)
==3691== Block was alloc'd at
==3691== at 0x4C2FB0F: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==3691== by 0x108C11: matrix_allocate (cdio.c:8)
==3691== by 0x108A80: main (del.c:29)
==3691==
83 86 77 15
49 21 62 27
90 59 63 26
40 26 72 36
==3741== Invalid read of size 8
==3741== at 0x108BA4: main (del.c:50)
==3741== Address 0x522d040 is 0 bytes inside a block of size 40 free'd
==3741== at 0x4C31D2F: realloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==3741== by 0x108A1C: del (del.c:12)
==3741== by 0x108B63: main (del.c:43)
==3741== Block was alloc'd at
==3741== at 0x4C2FB0F: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==3741== by 0x108C11: matrix_allocate (cdio.c:8)
==3741== by 0x108A80: main (del.c:29)
==3741==
==3741== Invalid free() / delete / delete[] / realloc()
==3741== at 0x4C30D3B: free (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==3741== by 0x108BCD: main (del.c:52)
==3741== Address 0x522d040 is 0 bytes inside a block of size 40 free'd
==3741== at 0x4C31D2F: realloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==3741== by 0x108A1C: del (del.c:12)
==3741== by 0x108B63: main (del.c:43)
==3741== Block was alloc'd at
==3741== at 0x4C2FB0F: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==3741== by 0x108C11: matrix_allocate (cdio.c:8)
==3741== by 0x108A80: main (del.c:29)
该函数输出一个实数数组。但这会导致不显着的零。
有没有简单的方法可以切断它们？
例如，有
2.00000 0.00000 421.12340 5.12909
它需要是：
2. 0. 421.1234 5.12909
请。
程序删除矩阵中的一行。一切正常，但无法从已删除的行中释放内存。
请告诉我，这怎么办？
#include "cdio.h"
#include "func.h"
int main()
{
long int **matrix = NULL;
size_t n = 5;
size_t m = 4;
matrix = matrix_allocate(n, m);
for (size_t i = 0; i < n; i++)
for (size_t j = 0; j < m; j++)
matrix[i][j] = rand() % 100;
puts("OLD MATRIX:");
print_matrix(matrix, n, m);
int pos = 0;
puts("enter position:");
scanf("%d", &pos);
for (size_t i = pos; i < n-1; i++)
matrix[i] = matrix[i+1];
n--;
long int **temp = realloc(matrix, n * sizeof(long int));
if (temp)
{
matrix = temp;
temp = NULL;
}
puts("NEW MATRIX:");
print_matrix(matrix, n, m);
for (int i = 0; i < n; i++)
free(matrix[i]);
free(matrix);
return 0;
}
==5203== HEAP SUMMARY:
==5203== in use at exit: 32 bytes in 1 blocks
==5203== total heap usage: 9 allocs, 8 frees, 2,280 bytes allocated
==5203==
==5203== LEAK SUMMARY:
==5203== definitely lost: 32 bytes in 1 blocks
==5203== indirectly lost: 0 bytes in 0 blocks
==5203== possibly lost: 0 bytes in 0 blocks
==5203== still reachable: 0 bytes in 0 blocks
==5203== suppressed: 0 bytes in 0 blocks
问题是什么？
// функция для удаления элемента массива
int remove_element(double *array, int *n, int pos)
{
if (pos >= *n || pos < 0)
{
printf("Incorrect position\n");
return ERROR_POSITION;
}
double *tmp = (double*)calloc(*n, sizeof(double));
if (!pos)
for (int i = 0; i < pos; i++)
tmp[i-1] = array[i];
else
{
for (int i = 0; i < pos; i++)
tmp[i] = array[i];
for (int i = pos; i < *n; i++)
tmp[i-1] = array[i];
}
free(array);
(*n)--;
array = (double*)malloc(*n * sizeof(double));
for (int i = 0; i < *n; i++)
array[i] = tmp[i];
free(tmp);
return SUCCESS;
}