int u32; typedef signed long long s64; typedef unsigned long long u64; 与体系结构相关的，定义在include/linux...int64_t; 对于各种数据类型的打印方式总结如下如下： 数据类型 打印格式 u8 %d s8 %d u16 %d or %hu s16 %d or %hd u32 %u s32 %d u64 %llu...d unsigned int %u short int %d or %hd long %ld unsigned long %lu long long %lld unsigned long long %llu...八进制 %0llo float %f double %f or %lf 科学技术类型(必须转化为double类型) %e 限制输出字段宽度 %x.yf (x:整数长度,y:小数点长度) 待解问题，在linux...definitions are present. */ #define __bool_true_false_are_defined 1 #endif /* stdbool.h */ 也大致解释了linux
技术背景我们在对接Linux平台RTSP播放模块的时候，遇到这样的技术需求，开发者需要把Linux RTSP播放器拉取的数据，除了实时播放外，还要投递给python，用于视觉算法分析。...技术实现Linux平台RTSP、RTMP直接播放不再赘述，这块我们非常成熟，python需要数据，我们可以在播放的同时，直接把数据回上来。...frame)return;fprintf(stdout, "OnSDKVideoFrameCallBack handle:%p frame:%p, timestamp:%llu\n", handle,.../outbitmaps/%llu.bmp", (unsigned long long)local_time_us);if (!...总结Linux平台RTSP、RTMP播放器数据跟python交互，两种方式均可，bitmap实现，也不麻烦，需要注意的时候，由于解码后的单帧数据比较大，建议适当控制导出的bitmap文件数。
%7llu %4llu %4llu %4llu %5llu %10llu %9llu " "%8llu %7llu %4llu %4llu %4llu %5llu %7llu %...说明丢包发生在Linux内核的缓冲区中。接下来，我们要继续探索到底是什么缓冲区引起了丢包问题，这就需要完整地了解服务器接收数据包的过程。...通知系统内核处理（驱动与Linux内核交互） 这个时候，数据包已经被转移到了sk_buffer中。...以下调用关系基于Linux-3.10.108及内核自带驱动drivers/net/ethernet/intel/ixgbe： ?.../** * 文件：include/linux/netdevice.h * napi_schedule - schedule NAPI poll * @n: NAPI context * *
mi.lpBaseOfDll); printf("Image Size: %u\n", (ULONG)mi.SizeOfImage); printf("Entry Point: 0x%llu...char*)malloc(8*sizeof(unsigned char*)); size_t nBytesRead; printf("Starting search from: 0x%llu...(LPVOID)(startAddress + currentOffset); if(DEBUG > 0){ printf("Searching at 0x%llu..."\n"); } if (memcmp(egg, current, 8) == 0) { printf("Found at %llu...int)-1), currentAddress, replace, 8, &nBytesRead); } } printf("Ended search at: 0x%llu
Tricks in Bits Time Limit: 1000MS Memory Limit: 65535KB 64bit IO Format: %lld & %llu Submit Status...cas=0; while(t--) { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%llu...MAX=1; MAX<<=63; ans=MAX; dfs(a[1],2); dfs(~a[1],2); printf("Case #%d: %llu
long result; result = (unsigned long long)(total_space_in_mb << 19); printf("wrong result:%llu...2952790016 */ result = ((unsigned long long)total_space_in_mb << 19); printf("right result:%llu...-S test.c可以生产名为test.s的汇编代码文件，内容如下： .file "test.c" .section .rodata .LC0: .string "wrong result:%llu...\n" .LC1: .string "right result:%llu\n" .text .globl main .type main, @function main: .LFB0: .cfi_startproc
sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); unsigned long long n, k; scanf("%llu...%llu", &n, &k); vector v(k + 4); unsigned long long t = 1; long long...t *= 10; } for (int i = 0; i < n - l; ++i) { printf("%d", 0); } printf("%llu...%llu", &n, &k); unsigned long long t = p(n); while(n > 0) { --n; unsigned long...%llu", &n, &k); k ^= (k / 2); while (n > 0) { --n; printf("%llu", k >> n & 1)
Case = 1; scanf("%d",&T); while(T--){ scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%llu...int j=i+1;j<n;j++){ ans = max(ans, fun(pre[i], pre[j])); } } printf("Case #%d: %llu
Ans *= a + 1; } printf("Between %llu...and %llu, %llu has a maximum of %u divisors...else { dfs(1, 1, 1); printf("Between %llu...and %llu, %llu has a maximum of %u divisors
[mid]-a[left]),min(a[right]-a[mid],l-a[right])); return maxx-minn; } signed main(){ scanf("%llu...%llu",&n,&l),n--; for(int i=1;i<=n;i++)scanf("%llu",&a[i]); if(a[n]!...1],l)<dif(a[mid],a[right],l))right++; ans=min(ans,calc(left,mid,right)); } printf("%llu
//[l,r]作为因数的贡献，相当于求一个等比数列的值 } return ans; } int main() { ll a,b; while((scanf("%llu...%llu", &a, &b)!
通过排查线上问题基本确定了是由于linux内核panic造成的原因，通过两个阶段的问题排查，基本上确定了linux内核panic的原因。...…..”的信息，但是根据相关信息，这个是不会导致linux系统挂起的。...%llu” 1980 “ ts=%llu write stamp = %llu\n%s”, 1981 (unsigned long long)*delta, 1982 ...Environment（环境） •Red Hat Enterprise Linux 6 ◦Red Hat Enterprise Linux 6.0, 6.1 and 6.2 are affected...•Red Hat Enterprise Linux 5 ◦architecture x86_64/64bit ■Red Hat Enterprise Linux 5.x: upgrade to
lld\n", g, h); unsigned long long i = 18446744073709551615; printf("unsigned long long可表示的范围：0 ~ %llu...308 15位 ---- 2.3 代码实例 #include #include int main() { printf("float 存储最大字节数 : %llu...printf("float 最大值: %E\n", FLT_MAX); printf("float 精度值: %d\n\n", FLT_DIG); printf("double 存储最大字节数 : %llu...double 最大值: %E\n", DBL_MAX); printf("double 精度值: %d\n\n", DBL_DIG); printf("long double 存储最大字节数 : %llu
printf("long long的范围为: %lld —— %lld\n", LLONG_MIN, LLONG_MAX); printf("unsigned long long的范围为: %llu...—— %llu\n\n", 0, ULLONG_MAX); return 0; } 程序输出结果为： 其中，CHAR_MIN、CHAR_MAX等符号常量是从哪来的呢？
bsr_int64_(unsigned __int64 num) { __int8 count=(sizeof(num)<<3)-1; for(unsigned __int64 mask=1LLU...return count; } 基本的思路就是用for循环从最高位开始对每一位做与运算，找到第一个为1的位，就中止循环，count中就是结果，如果所有的位都为0，则count为-1; 注意这里1LLU...<< count, LLU限定前面的数字1为long long(64位)，U限定为无符号类型(unsigned)，就是指1为unsigned long long 型64位无符号整数。...bsr_int64_(unsigned __int64 num) { __int8 count=(sizeof(num)<<3)-1; for(unsigned __int64 mask=1LLU..._bsr(88080LLU);//调用static __int8 _bsr(unsigned __int64 num) 看完这些内容，如果你想实现bsf(正向位扫描)指令，也可以在这个基础上如法炮制了。
Flag format :- enclose entire thing in shellctf{} A - Thing you got after solving problem B - Linux command...me to flag then special command is “ls” and argument is “ls”) C,D,.. - Numbers of arguments use with Linux...Example flag - Suppose I got string “H3re_1s_F1ag” from solving so my A = H3re_1s_F1ag Suppose I used Linux...图片中存在LSB隐写，需要补俩等号 是带密码的7z文件，密码为之前brainfuck解码出来的名字里Luc4r10，解压得到part2_p4raM37eR_P0llu7iOn。...shellctf{USSERAGENT_p4raM37eR_P0llu7iOn}
typedef short unsigned int hu; typedef unsigned int uint; typedef long long unsigned int llu...static inline const bool isprime(const llu a) { if (a == 0 or a == 1) return false; for (register...static inline const bool isprime(const llu a) {...} enum { N = 1u << 20 }; static uint n; static bool...static uint n; static bool isp[N]; static uint prime[N], cp; static inline const bool isprime(const llu
/=MSECOND; printf("main thread:%x,Used Time:%f/n",pthread_self(),timeuse); printf("a = %llu...,b = %llu,sum=%llu/n",test.a,test.b,sum); return 0; } #ifdef __cplusplus } #endif 假设有一台酷睿
unsigned short a1=10; unsigned long a2 = 10lu; //简写成unsigned long a2=10; unsigned long long a3 = 10llu...这里打印short短整型要用hu printf("short a1=%hu\n", a1); printf("long a2=%lu\n", a2); printf("long long a3=%llu
tv_end.tv_sec - tv_begin.tv_sec) * 1000000 + (tv_end.tv_usec - tv_begin.tv_usec); printf("cost %llu...tv_end.tv_sec - tv_begin.tv_sec) * 1000000 + (tv_end.tv_usec - tv_begin.tv_usec); printf("cost %llu
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