我是Ubuntu和Git的新手。如何验证我是否在我的机器上安装了git?当我尝试的时候我得到了:
root@ubuntu:/home/nebojsa# apt-get install git
Reading package lists... Done
Building dependency tree
Reading state information... Done
git is already the newest version.
The following packages were automatically installed and are no longer req
我正在编译Linux4.19(gcc-8.2 bintutils-2.31),但是它总是失败,错误如下:
aarch64-oe-linux-ld.bfd: drivers/platform/gsi/gsi.o: relocation R_AARCH64_ABS32 against `__crc_gsi_write_channel_scratch' can not be used when making a shared object
aarch64-oe-linux-ld.bfd: drivers/platform/gsi/gsi.o:/usr/src/kernel/drivers/p
如果程序使用ptrace来检测它是否在调试器中运行,我发现了下面的技巧:
catch syscall ptrace
commands 1
set ($eax) = 0
continue
end
有人能解释一下它是怎么工作的吗?我试图在i r eax之后插入commands 1,但是我不理解我得到的负值:
Catchpoint 1 (call to syscall ptrace), 0x00007ffff778af1e in ptrace (request=PTRACE_TRACEME) at ../sysdeps/unix/sysv/linux/ptrace.c:45
45 ../sysde
我正在尝试定义一些在其中调用printf的子例程。下面是一个非常简单的例子:
extern printf
LINUX equ 80H
EXIT equ 60
section .data
intfmt: db "%ld", 10, 0
segment .text
global main
main:
call os_return ; return to operating system
os_return:
mov rax, EXIT ; Linux system call
对于相同的cacert.pem,openssl返回不同的结果。
Redhat linux 8证书验证失败,Redhat linux 7.5证书验证成功
[rhel8]: openssl verify cacert.pem
C = NO, L = Asker, O = E1, OU = ETO TER, CN = ETO Opto Certificate Authority
error 18 at 0 depth lookup: self signed certificate
error cacert.pem: verification failed
RHEL 7.5:
[rhel7.5]: o
以下是从Hibernate生成的查询(除了我将字段列表替换为*):
select *
from
resource resource0_,
resourceOrganization resourceor1_
where
resource0_.active=1
and resource0_.published=1
and (
resource0_.resourcePublic=1
or resourceor1_.resource_id=resource0_.id
and resourceor1_.or
我有一个关于openSSL库的问题,我有内存泄漏:
==9904==
==9904== HEAP SUMMARY:
==9904== in use at exit: 163,350 bytes in 2,867 blocks
==9904== total heap usage: 5,104 allocs, 2,237 frees, 1,162,949 bytes allocated
==9904==
==9904== 4,808 (184 direct, 4,624 indirect) bytes in 1 blocks are definitely lost in loss
当我在windows和Linux上运行下面写的代码时,我得到了两个不同的输出。
这两个我都用的是gcc。当我在windows上运行它时,我得到的输出是"Seek“,而在Linux上运行它时,我得到的输出是"Hide”。Windows和Linux的内存布局有什么不同,还是有其他原因导致输出不同?
int main()
{
int a=0;
int *b=(int *)malloc(sizeof(int));
if(&a>b)
printf("Hide");
else
printf(
我一直在阅读“外壳编码器手册”,并参考链接了解堆栈溢出的实践。但是看起来Linux内核开发者已经让内核变得非常安全。这是我的问题。
1)此代码
void function(int a, int b, int c) {
char buffer1[8];
char buffer2[10];
int* ret;
ret = buffer1 + 6;
*ret+=8;
}
void main() {
int x;
x = 0;
function(1,2,3);
x = 1;
printf("%d\n",x);
}
给出输出
$ c
为了保持代码的整洁(并且像我一直做的那样过于复杂),我希望main()函数将一个/pointer传递给int/数组。然后,该函数将按程序展开数组,并添加符合条件的新条目。
我有一个可以工作的代码,但是由于某种原因,valgrind报告了一个内存泄漏(来自主数组)。而且,我不喜欢在重新分配数组之前分配数组的方式。
#include <iostream>
#include <cstdlib>
int get_divisors(int num_input, int *num_result);
int main()
{
int num_input, *num_res
我有一个生产VMU16.04LTS,它打破了apt和完整的/boot目录与各种内核。
apt-get install -f
Reading package lists... Done
Building dependency tree
Reading state information... Done
Correcting dependencies... Done
The following packages were automatically installed and are no longer required:
linux-headers-4.4.0-138 lin