memcpy

I am doing Opengl-es 2.0 in ununtu 10.10 through the use of kronos and pvrsdk .Now code #include <GLES2/gl2.h> #include <GLES2/gl2ext.h> ==========|||||||||||||||||||||||||||||||||||=================== GLfloat *pData = glMapBufferOES (GL_ARRAY_BUFFER, GL_WRITE_ONLY_OES); for(i=0; i<triNum[surfnum]; ++i,pData+=9) { memcpy(pData, triArray[surfnum][i].pt1, 3*sizeof(GLfloat)); memcpy(pData+3, triArray[surfnum][i].pt2, 3*sizeof(GLfloat)); memcpy(pData+6, triArray[surfnum][i].pt3, 3*sizeof(GLfloat)); } glUnmapBufferOES (GL_ARRAY_BUFFER); Error : src/Hello.cpp: In function 'int main(int, char**)':

strcpy和memcpy都是标准C库函数，它们有下面的特点。 strcpy提供了字符串的复制。即strcpy只用于字符串复制，并且它不仅复制字符串内容之外，还会复制字符串的结束符。 已知strcpy函数的原型是：char* strcpy(char* dest, const char* src); memcpy提供了一般内存的复制。即memcpy对于需要复制的内容没有限制，因此用途更广。 void *memcpy( void * dest , const void * src , size_t count ); char * strcpy ( char * dest, const char * src) // 实现src到dest的复制 { 　　 if ((src == NULL) || (dest == NULL)) //判断参数src和dest的有效性 　　{ 　　　　　　 return NULL; 　　} 　　 char *strdest = dest; //保存目标字符串的首地址 　　 while ((*strDest++ = *strSrc++)!= '\0' ); //把src字符串的内容复制到dest下 　　 return strdest; } void * memcpy ( void *memTo, const void *memFrom, size_t size)

We are dealing with C here. I'm just had this idea, wondering if it is possible to access the point in memory where a function is stored, say foo and copying the contents of the function to another point in memory. Specifically, I'm trying to get the following to work: #include <stdlib.h> #include <stdio.h> #include <string.h> void foo(){ printf("Hello World"); } int main(){ void (*bar)(void) = malloc(sizeof foo); memcpy(&bar, &foo, sizeof foo); bar(); return 0; } But running it gives a bus error: Bus error: 10 . I'm trying to copy over the contents of function foo into a space of memory bar

I would like to compare the GCC builtin function memcpy versus the one one from libc. However, all iterations of -fno-builtin or -fno-builtin-memcpy seem to be ignored. //g++ -O3 foo.cpp -S or //g++ -O3 -fno-builtin foo.cpp -S #include <string.h> int main() { volatile int n = 1000; //int n = 1000; float *x = new float[1000]; float *y = new float[1000]; memcpy(y,x,sizeof(float)*n); //__builtin_memcpy(y,x,sizeof(float)*n); } What I have found is that if n in the source code above is not volatile then it inlines built-in code. However, when n is made volatile then it calls the function __memcpy

C++ 风格的复制操作 使用STL中的copy算法 int a[] = {1,2,3,4,5}; int b[5]; std::copy(std::begin(a),std::end(a),std::begin(b)); for(auto e:b) cout<<e<<" "; // 输出 1,2,3,4,5 上述程序中，copy算法将数组a区间中的数复制到以begin(b)开始的区间中去. 使用array容器 (C++11) std::array<int,5> arr = {1,2,3,4,5}; std::array<int,5> copy; copy = arr; // 将arr中的元素复制到copy中 arr[0] = 100; for(auto e:copy) cout<<e<<" "; //输出 1,2,3,4,5 C 风格的复制操作 使用memcpy() int arr[] = {1,2,3,4,5}; int copy[5]; int len = sizeof(arr) / sizeof(arr[0]); memcpy(copy,arr,len*sizeof(int)); // 输出 1,2,3,4,5 for(auto e:copy) cout<<e<<" "; 注意：memcpy()函数的第三个参数表示的是要复制的字节数，而不是要复制的元素数目。至于这样做的原因

Say i have an array in C int array[6] = {1,2,3,4,5,6} how could I split this into {1,2,3} and {4,5,6} Would this be possible using memcpy? Thank You, nonono Sure. The straightforward solution is to allocate two new arrays using malloc and then using memcpy to copy the data into the two arrays. int array[6] = {1,2,3,4,5,6} int *firstHalf = malloc(3 * sizeof(int)); if (!firstHalf) { /* handle error */ } int *secondHalf = malloc(3 * sizeof(int)); if (!secondHalf) { /* handle error */ } memcpy(firstHalf, array, 3 * sizeof(int)); memcpy(secondHalf, array + 3, 3 * sizeof(int)); However, in case the