Program: #include<stdio.h> #include<sys/types.h> #include<malloc.h> main() { int *i1, *i2; printf("sbrk(0) before malloc(): %xn", sbrk(0)); i1 = (int *) malloc(sizeof(int)); printf("sbrk(0) after `i1 = (int *) malloc(4)': %xn", sbrk(0)); i2 = (int *) malloc(sizeof(int)); printf("sbrk(0) after `i2 = (int *) malloc(4)': %xn", sbrk(0)); } Output 1: mohan
程序： #include<stdio.h> #include<sys/types.h> #include<malloc.h> main() { int *i1, *i2; printf("sbrk(0) before malloc(): %xn", sbrk(0)); i1 = (int *) malloc(sizeof(int)); printf("sbrk(0) after `i1 = (int *) malloc(4)': %xn", sbrk(0)); i2 = (int *) malloc(sizeof(int)); printf("sbrk(0) after `i2 = (int *) malloc(4)': %xn", sbrk(0)); } 输出1： mohanraj@
Program: #include<stdio.h> #include<sys/types.h> #include<malloc.h> main() { int *i1, *i2; char *s; printf("sbrk(0) before malloc(4): %xn", sbrk(0)); i1 = (int *) malloc(sizeof(int)); printf("sbrk(0) after `i1 = (int *) malloc(4)': %xn", sbrk(0)); i2 = (int *) malloc(sizeof(int)); printf("sbrk(0) after `i2 = (int *) malloc(4)': %xn", sbrk(0)); } Out
程序： #include<stdio.h> #include<sys/types.h> #include<malloc.h> main() { int *i1, *i2; char *s; printf("sbrk(0) before malloc(4): %xn", sbrk(0)); i1 = (int *) malloc(sizeof(int)); printf("sbrk(0) after `i1 = (int *) malloc(4)': %xn", sbrk(0)); i2 = (int *) malloc(sizeof(int)); printf("sbrk(0) after `i2 = (int *) malloc(4)': %xn", sbrk(0)); } 输出
I was trying to figure out how much memory I can malloc to maximum extent on my machine (1 Gb RAM 160 Gb HD Windows platform). I read that the maximum memory malloc can allocate is limited to physical memory (on heap). Also when a program exceeds consumption of memory to a certain level, the computer stops working because other applications do not get enough memory that they require. So to
我试图弄清楚我的机器上最多可以使用多少内存（1 Gb RAM 160 Gb HD Windows平台）。 我读了malloc可以分配的最大内存限于物理内存（在堆上）。 另外，当某个程序将内存消耗量超过一定水平时，计算机将停止工作，因为其他应用程序没有获得足够的内存。 为了确认，我用C写了一个小程序： int main(){ int *p; while(1){ p=(int *)malloc(4); if(!p)break; } } 我希望有一段时间内存分配
Using a function like this: #include <stdio.h> #include <stdlib.h> #include <sys/wait.h> #include <unistd.h> void print_trace() { char pid_buf[30]; sprintf(pid_buf, "--pid=%d", getpid()); char name_buf[512]; name_buf[readlink("/proc/self/exe", name_buf, 511)]=0; int child_pid = fork(); if (!child_pid) { dup2(2,1); // redirect ou
使用这样的函数： #include <stdio.h> #include <stdlib.h> #include <sys/wait.h> #include <unistd.h> void print_trace() { char pid_buf[30]; sprintf(pid_buf, "--pid=%d", getpid()); char name_buf[512]; name_buf[readlink("/proc/self/exe", name_buf, 511)]=0; int child_pid = fork(); if (!child_pid) { dup2(2,1); // redirect output to std
Problem I am learning about HPC and code optimization. I attempt to replicate the results in Goto's seminal matrix multiplication paper (http://www.cs.utexas.edu/users/pingali/CS378/2008sp/papers/gotoPaper.pdf). Despite my best efforts, I cannot get over ~50% maximum theoretical CPU performance. Background See related issues here (Optimized 2x2 matrix multiplication: Slow assembly ver
问题 我正在学习HPC和代码优化。 我试图在Goto的开创性矩阵乘法论文中复制结果（http://www.cs.utexas.edu/users/pingali/CS378/2008sp/papers/gotoPaper.pdf）。 尽管我尽了最大的努力，但我无法超过理论CPU性能的50％。 背景 在此处查看相关问题（优化的2×2矩阵乘法：组装速度较慢，相对于快速SIMD），包括有关我的硬件的信息 我所尝试过的 这篇相关论文（http://www.cs.utexas.edu/users/flame/pubs/blis3_ipdps1
This question already has an answer here: Why is transposing a matrix of 512x512 much slower than transposing a matrix of 513x513? 3 answers Why is my program slow when looping over exactly 8192 elements? 3 answers
这个问题在这里已经有了答案： 为什么转置一个512x512的矩阵要比转置513x513的矩阵慢得多？ 3个答案 为什么我的程序在循环8192个元素时很慢？ 3个答案
I found this bit operation in a source code: A = 0b0001; B = 0b0010; C = 0b0100; flags |= !!(flags & (A | B)) * C; I can't see, why this complicated expression is used. flags & (A | B) filters flags to A | B A | B . Now it's converted to true , if flags is set to anything, and false otherwise. true * C == C and false * C == 0 . Is it slower to just use flags = flags ? flag
我在源代码中发现了这一点操作： A = 0b0001; B = 0b0010; C = 0b0100; flags |= !!(flags & (A | B)) * C; 我看不出，为什么使用这个复杂的表达式。 flags & (A | B) flags过滤到A | B A | B 现在，如果flags设置为任何true ，则将其转换为true ，否则为false 。 true * C == C和false * C == 0 。 仅仅使用flags = flags ? flags | C会慢flags = flags ? flags | C flags = flags ? flags | C flags = flags
Systems demand that certain primitives be aligned to certain points within the memory (ints to bytes that are multiples of 4, shorts to bytes that are multiples of 2, etc.). Of course, these can be optimized to waste the least space in padding. My question is why doesn't GCC do this automatically? Is the more obvious heuristic (order variables from biggest size requirement to smallest) la
系统要求某些原语与内存中的某些点对齐（整数字节是4的倍数，短到字节的倍数是2等）。 当然，这些可以进行优化，以减少填充中的最小空间。 我的问题是为什么GCC不会自动执行此操作？ 更明显的启发式（从最大尺寸要求到最小的顺序变量）缺乏某种方式？ 是否有一些代码依赖于它的结构的物理顺序（这是一个好主意）？ 我只是问，因为海湾合作委员会是在很多方面超级优化，但没有在这一个，我认为必须有一些相对较酷的解释（
If you check this very nice page: http://www.codeproject.com/Articles/69941/Best-Square-Root-Method-Algorithm-Function-Precisi You'll see this program: #define SQRT_MAGIC_F 0x5f3759df float sqrt2(const float x) { const float xhalf = 0.5f*x; union // get bits for floating value { float x; int i; } u; u.x = x; u.i = SQRT_MAGIC_F - (u.i >> 1); // gives initial
如果你检查这个非常好的页面： http://www.codeproject.com/Articles/69941/Best-Square-Root-Method-Algorithm-Function-Precisi 你会看到这个程序： #define SQRT_MAGIC_F 0x5f3759df float sqrt2(const float x) { const float xhalf = 0.5f*x; union // get bits for floating value { float x; int i; } u; u.x = x; u.i = SQRT_MAGIC_F - (u.i >> 1); // gives initial guess y0 ret
Addition mathematically holds the associative property: (a + b) + c = a + (b + c) In the general case, this property does not hold for floating-point numbers because they represent values in a finite precision. Is a compiler allowed to make the above substitution when generating machine code from a C program as part of an optimization? Where does it exactly say in the C standard? The compi
加法数学上保持关联属性： (a + b) + c = a + (b + c) 在一般情况下，该属性不适用于浮点数，因为它们以有限精度表示值。 作为优化的一部分，是否允许编译器在从C程序生成机器代码时进行上述替换？ C标准究竟在哪里说的？ 编译器不允许执行“优化”，这会导致计算出不同的值，而不是根据抽象机器语义计算的值。 5.1.2.3程序执行 [＃1]本国际标准中的语义描述描述了优化问题无关的抽象机器的行为。 [＃3]在抽象机器中