为什么这段代码不安全?
#include <stdio.h>
int main( int argc, char *argv[] )
{
printf(argv[1]);
printf("\n");
return 0;
}
Run Code Online (Sandbox Code Playgroud) 我有一个名为的内核模块mymodule,我有:
static struct file_operations my_mod_fops = {
.owner = THIS_MODULE
.write = my_write,
.open = my_open,
.unlocked_ioctl = my_ioctl,
.read = my_read,
};
Run Code Online (Sandbox Code Playgroud)
测试了前一个结构中映射的所有函数,它们都没问题.我想声明全局(并且静态地,如果可用)一个char kernel_array[128],我想编写一个可以执行此操作的用户空间应用程序:
int main(){
char* ptr_to_kernel_arr = get_kernel_array_address();
for (int i=0 ; i<128;++i)
*(ptr_to_kernel_arr+i) = i;
return 0;
}
Run Code Online (Sandbox Code Playgroud)
我的困难是:
如何获取kernel_array[128]可以从用户空间应用程序分配值的地址?
内核如何知道其lsmod列表中的哪个模块拥有该模块kernel_array
如何处理mmap所有这些情况?
我想知道,是否可以定义一个API并将其作为TM图灵机的输入,输出将是代码c或任何其他自然/编程语言?
我猜不是,但我如何通过减少等正式展示?
$ sudo pip install beautifulsoup4
Requirement already satisfied (use --upgrade to upgrade): beautifulsoup4 in /Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages
Cleaning up...
Run Code Online (Sandbox Code Playgroud)
我已安装beautifulsoup4并且似乎已成功完成但我无法导入它:
Python 2.7.3 (v2.7.3:70, Apr 9 2012, 20:52:43)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import beautifulsoup4
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ImportError: No module named beautifulsoup4
>>> import beautifulsoup
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ImportError: …Run Code Online (Sandbox Code Playgroud) 我想在c中写一个python扩展.我在Mac上工作,我从这里拿了代码:
#include <Python.h>
static PyObject* say_hello(PyObject* self, PyObject* args)
{
const char* name;
if (!PyArg_ParseTuple(args, "s", &name))
return NULL;
printf("Hello %s!\n", name);
Py_RETURN_NONE;
}
static PyMethodDef HelloMethods[] =
{
{"say_hello", say_hello, METH_VARARGS, "Greet somebody."},
{NULL, NULL, 0, NULL}
};
PyMODINIT_FUNC
inithello(void)
{
(void) Py_InitModule("hello", HelloMethods);
}
Run Code Online (Sandbox Code Playgroud)
我编译它:
gcc -c -o py_module.o py_module.c -I/Library/Frameworks/Python.framework/Versions/2.7/include/python2.7/
gcc -o py_module py_module.o -I/Library/Frameworks/Python.framework/Versions/2.7/include/python2.7/ -lm
Run Code Online (Sandbox Code Playgroud)
但我得到这个错误:
Undefined symbols for architecture x86_64:
"_PyArg_ParseTuple", referenced from:
_say_hello in py_module.o
"_Py_InitModule4_64", referenced from:
_inithello in py_module.o
"__Py_NoneStruct", …Run Code Online (Sandbox Code Playgroud) 我正在研究一些将RGBA图像转换为灰度的cuda 教程.但我无法弄清楚为什么改变它blockSize并gridSize进行X33时间的改进.
__global__
void rgba_to_greyscale(const uchar4* const rgbaImage,
unsigned char* const greyImage,
int numRows, int numCols)
{
int i = blockIdx.x*numCols + threadIdx.x;
float channelSum = .299f * rgbaImage[i].x + .587f * rgbaImage[i].y + .114f * rgbaImage[i].z;
greyImage[i]= channelSum;
}
void your_rgba_to_greyscale(const uchar4 * const h_rgbaImage, uchar4 * const d_rgbaImage,
unsigned char* const d_greyImage, size_t numRows, size_t numCols)
{
const dim3 blockSize(numCols, 1, 1);
const dim3 gridSize(numRows, 1 , 1);
rgba_to_greyscale<<<gridSize, blockSize>>>(d_rgbaImage, d_greyImage, numRows, numCols); …Run Code Online (Sandbox Code Playgroud) 我担心我误解了VBA的excel文档,我有这条似乎是一个错误的行:
Range a = Selection.SpecialCells(xlCellTypeConstants, 23)
Run Code Online (Sandbox Code Playgroud)
但是这个很好:
Set a = Selection.SpecialCells(xlCellTypeConstants, 23)
Run Code Online (Sandbox Code Playgroud)
文件声称:
返回一个Range 对象,该对象表示与指定类型和值匹配的所有单元格.
但它实际上返回一个byRef对象,这就是我必须使用的原因Set.
我在这里想念什么?
这是Range.SpecialCellsExcel中的方法帮助:

任何想法为什么我在尝试导入时会出现此错误ctypes:
>>> from ctypes import *
Traceback (most recent call last):
File "<pyshell#4>", line 1, in <module>
from ctypes import *
File "C:/Python27\ctypes.py", line 3, in <module>
libc = ctypes.windll.msvcrt
AttributeError: 'module' object has no attribute 'windll'
Run Code Online (Sandbox Code Playgroud)