Search Results

Search found 138 results on 6 pages for 'vaibhav gade'.

Page 6/6 | < Previous Page | 2 3 4 5 6 

  • Difference between a Deprecated and an Legacy API?

    - by Vaibhav Bajpai
    I was studying the legacy API's in the Java's Collection Framework and I learnt that classes such as Vector and HashTable have been superseded by ArrayList and HashMap. However still they are NOT deprecated, and deemed as legacy when essentially, deprecation is applied to software features that are superseded and should be avoided, so, I am not sure when is a API deemed legacy and when it is deprecated.

    Read the article

  • Usage of Native in Java.

    - by vaibhav
    I am basically a web application developer, never in my career got a chance of trying out this.. can someone explain the whys and the wherefores for using Native code in java? Thanks a Ton in advance!

    Read the article

  • CUDA memory transfer issue

    - by Vaibhav Sundriyal
    I am trying to execute a code which first transfers data from CPU to GPU memory and vice-versa. In spite of increasing the volume of data, the data transfer time remains the same as if no data transfer is actually taking place. I am posting the code. #include <stdio.h> /* Core input/output operations */ #include <stdlib.h> /* Conversions, random numbers, memory allocation, etc. */ #include <math.h> /* Common mathematical functions */ #include <time.h> /* Converting between various date/time formats */ #include <cuda.h> /* CUDA related stuff */ #include <sys/time.h> __global__ void device_volume(float *x_d,float *y_d) { int index = blockIdx.x * blockDim.x + threadIdx.x; } int main(void) { float *x_h,*y_h,*x_d,*y_d,*z_h,*z_d; long long size=9999999; long long nbytes=size*sizeof(float); timeval t1,t2; double et; x_h=(float*)malloc(nbytes); y_h=(float*)malloc(nbytes); z_h=(float*)malloc(nbytes); cudaMalloc((void **)&x_d,size*sizeof(float)); cudaMalloc((void **)&y_d,size*sizeof(float)); cudaMalloc((void **)&z_d,size*sizeof(float)); gettimeofday(&t1,NULL); cudaMemcpy(x_d, x_h, nbytes, cudaMemcpyHostToDevice); cudaMemcpy(y_d, y_h, nbytes, cudaMemcpyHostToDevice); cudaMemcpy(z_d, z_h, nbytes, cudaMemcpyHostToDevice); gettimeofday(&t2,NULL); et = (t2.tv_sec - t1.tv_sec) * 1000.0; // sec to ms et += (t2.tv_usec - t1.tv_usec) / 1000.0; // us to ms printf("\n %ld\t\t%f\t\t",nbytes,et); et=0.0; //printf("%f %d\n",seconds,CLOCKS_PER_SEC); // launch a kernel with a single thread to greet from the device //device_volume<<<1,1>>>(x_d,y_d); gettimeofday(&t1,NULL); cudaMemcpy(x_h, x_d, nbytes, cudaMemcpyDeviceToHost); cudaMemcpy(y_h, y_d, nbytes, cudaMemcpyDeviceToHost); cudaMemcpy(z_h, z_d, nbytes, cudaMemcpyDeviceToHost); gettimeofday(&t2,NULL); et = (t2.tv_sec - t1.tv_sec) * 1000.0; // sec to ms et += (t2.tv_usec - t1.tv_usec) / 1000.0; // us to ms printf("%f\n",et); cudaFree(x_d); cudaFree(y_d); cudaFree(z_d); return 0; } Can anybody help me with this issue? Thanks

    Read the article

  • convert pointer to pointer to void pointer

    - by FihopZz
    When I'm learning to use qsort to sort an array of string, there is a question puzzled me. For example, to sort the following s char *s[] = { "Amit", "Garima", "Gaurav", "Vaibhav" }; To use the qsort, you must provide a comparison function like the following function cstring_cmp I guess in the qsort function, the type of parameter to be passed to the function cstring_cmp is char**. How to convert a char** to a void*? Why can we convert a char** to a void*? int cstring_cmp(const void *a, const void *b) { const char **ia = (const char **)a; const char **ib = (const char **)b; return -strcasecmp(*ia, *ib); /* return the negative of the normal comparison */ }

    Read the article

< Previous Page | 2 3 4 5 6