我正在开发一种模式，如果它不包含两个匹配的＃，则验证字符串。这是代码：
^[^\!|\@|\$|\%|\^|\&|\*|\+][\w]*([\w ]+\#[\w ]*){0,2}$
[！^ | \ @ | \ $ | \ | \ ^ | \＆安培; | * | +]
这是一组不可接受的符号。
此外，如果字符串包含其他符号（ - _，。/），则该模式应验证字符串。每个符号都应该有自己的计数器，不应该在任何位置匹配两次以上。
例如，如果我有这样的字符串：
Mcloud博士。 ＃33 /＃45，一些文字，一些文字
它应该是有效的。但在这种情况下不应该：
Mcloud博士。 ### 33 / $＃45，####，----
你会建议什么？
答案 0 :(得分：1)
鉴于你想匹配字母数字字符和一些特殊符号#include <device_launch_parameters.h> // threadIdx
#include <stdio.h> // fprintf
#include <math.h> // ceil
#include "simple_math.cuh"
cudaError_t mathWithCuda(double *c, const float *a, const float *b, unsigned int arraySize, int mode)
{
float *dev_a, *dev_b;
double *dev_c;
cudaError_t cudaStatus;
cudaStatus = cudaSetDevice(0);
if (cudaStatus != cudaSuccess) {
fprintf(stderr, "cudaSetDevice failed! Do you have a CUDA-capable GPU installed?");
goto Error;
}
if ((cudaStatus = cudaMalloc((void**)&dev_c, arraySize * sizeof(double))) != cudaSuccess ||
(cudaStatus = cudaMalloc((void**)&dev_a, arraySize * sizeof(float))) != cudaSuccess ||
(cudaStatus = cudaMalloc((void**)&dev_b, arraySize * sizeof(float))) != cudaSuccess)
{
fprintf(stderr, "cudaMalloc failed!");
goto Error;
}
if ((cudaStatus = cudaMemcpy(dev_a, a, arraySize * sizeof(float), cudaMemcpyHostToDevice)) != cudaSuccess ||
(cudaStatus = cudaMemcpy(dev_b, b, arraySize * sizeof(float), cudaMemcpyHostToDevice)) != cudaSuccess)
{
fprintf(stderr, "cudaMemcpy failed!");
goto Error;
}
int blocksPerGrid, threadsPerBlock;
if (arraySize < 512) {
blocksPerGrid = 1;
threadsPerBlock = arraySize;
}
else {
blocksPerGrid = ceil(double(arraySize) / double(threadsPerBlock));
threadsPerBlock = 512;
}
switch (mode)
{
case 0:
addKernel <<<blocksPerGrid, threadsPerBlock >>>(dev_c, dev_a, dev_b);
break;
case 1:
subKernel <<<blocksPerGrid, threadsPerBlock >>>(dev_c, dev_a, dev_b);
break;
case 2:
mulKernel <<<blocksPerGrid, threadsPerBlock >>>(dev_c, dev_a, dev_b);
break;
case 3:
divKernel <<<blocksPerGrid, threadsPerBlock >>>(dev_c, dev_a, dev_b);
break;
default:
// nothing
break;
}
cudaStatus = cudaGetLastError();
if (cudaStatus != cudaSuccess) {
fprintf(stderr, "Kernel launch failed: %s\n", cudaGetErrorString(cudaStatus));
goto Error;
}
cudaStatus = cudaDeviceSynchronize();
if (cudaStatus != cudaSuccess) {
fprintf(stderr, "cudaDeviceSynchronize returned error code %d after launching Kernel!\n", cudaStatus);
goto Error;
}
cudaStatus = cudaMemcpy(c, dev_c, arraySize * sizeof(float), cudaMemcpyDeviceToHost);
if (cudaStatus != cudaSuccess) {
fprintf(stderr, "cudaMemcpy failed!");
goto Error;
}
Error:
cudaFree(dev_c);
cudaFree(dev_a);
cudaFree(dev_b);
return cudaStatus;
}
__global__ void addKernel(double *c, const float *a, const float *b)
{
int i = threadIdx.x;
c[i] = __fadd_rn(a[i], b[i]); // a + b
}
__global__ void subKernel(double *c, const float *a, const float *b)
{
int i = threadIdx.x;
c[i] = __fsub_rn(a[i], b[i]); // a - b
}
__global__ void mulKernel(double *c, const float *a, const float *b)
{
int i = threadIdx.x;
c[i] = __fmul_rn(a[i], b[i]); // a * b
}
__global__ void divKernel(double *c, const float *a, const float *b)
{
int i = threadIdx.x;
c[i] = __fdividef(a[i], b[i]); // a/b
}
你必须在像这样的字符类中提及它们。
正则表达式： ()-_,./
<强>解释
^(?!.*([(),.#/-])\1)([\w (),.#/-]+)$断言不应该在字符类中提到多个字符。这从字符串的开头断言。
(?!.*([(),.#/-])\1)匹配字符串的其余部分，从头到尾包含允许的字符。