GDAL C++ RasterIO的块

问题描述：

我是gdal和C++的新手，我试图创建一个按块（使用RasterIO而不是ReadBlock）读取栅格的函数，并将结果写入新的栅格文件GDAL C++ RasterIO的块

我得到一个错误：

Unhandled exception at 0x00007FFDFB517BC4 (gdal201.dll) in Test.exe: 0xC0000005: Access violation writing location 0x00000230E2593000 at line rasterBand->RasterIO(GDALRWFlag::GF_Read, j * readCols, i * readRows, readCols, readRows, rasterBlock, readCols, readRows, GDALDataType::GDT_CFloat32, 0, 0)

我有一个非常类似的功能工作正常，但按行

任何帮助表示赞赏读取的像素值排

功能是波纹管

谢谢

#include <iostream> 
#include "gdal_priv.h" 
using namespace std; 

void SlopeBlock() 
{ 
int readRows = 2048; 
int readCols = 2048; 
int nBlockXSize; 
int nBlockYSize; 
int rowOff = 0; 
int colOff = 0; 
int nRows = 0; 
int nCols = 0; 
double noData = -9999; 

GDALAllRegister(); 

GDALDataset* dem = (GDALDataset*)GDALOpen("path/EUD_CP-DEMS_4500025000-AA.tif", GA_ReadOnly); 

GDALDriver *gTIF = GetGDALDriverManager()->GetDriverByName(dem->GetDriverName()); 

double geoTransform[6]; 
dem->GetGeoTransform(geoTransform); 

GDALRasterBand* rasterBand = dem->GetRasterBand(1); 

nCols = rasterBand->GetXSize(); 
nRows = rasterBand->GetYSize(); 
noData = rasterBand->GetNoDataValue(); 

//rasterBand->GetBlockSize(&nBlockXSize, &nBlockYSize); 

int nXBlocks = (rasterBand->GetXSize() + readCols - 1)/readCols; 
int nYBlocks = (rasterBand->GetYSize() + readRows - 1)/readRows; 

cout << nYBlocks << "___" << nXBlocks << "\n"; 


//Slope 
GDALDataset *slope = gTIF->Create("path/slopeBlock.tif", nCols, nRows, 1, GDT_Float32, NULL); 
slope->SetGeoTransform(geoTransform); 

for (int i = 0; i < nYBlocks; i++) 
{ 
    for (int j = 0; j < nXBlocks; j++) 
    { 
     float* rasterBlock = (float*)CPLMalloc(readCols*readRows*sizeof(float)); 
     float* rasterBlockOut = (float*)CPLMalloc(readCols*readRows*sizeof(float)); 

     if (rasterBand != nullptr) 
      rasterBand->RasterIO(GDALRWFlag::GF_Read, j * readCols, i * readRows, readCols, readRows, rasterBlock, readCols, readRows, GDALDataType::GDT_CFloat32, 0, 0); 

     //Not sure if is working 
     for (int jb = 0; jb < readCols; jb++) 
     { 
      if (rasterBlock[jb] == noData) 
      { 
       rasterBlockOut[jb] = noData; 
      } 
      else 
      { 
       rasterBlockOut[jb] = rasterBlock[jb]; 
      } 
     } 

     slope->GetRasterBand(1)->RasterIO(GDALRWFlag::GF_Write, j * readCols, i * readRows, readCols, readRows, rasterBlockOut, readCols, readRows, GDALDataType::GDT_CFloat32, 0, 0); 

     CPLFree(rasterBlock); 
     CPLFree(rasterBlockOut); 
    } 
} 

GDALClose(dem); 
GDALClose(slope);

}

答

这可能不是飞机坠毁的原因，但你问一个2048 * 2048块就算你的形象是较小。此代码只适用于尺寸为2048的倍数的图像。

你应该这样做：

nXSize = min(readCols, rasterBand->GetXSize() - j * readCols); 
nYSize = min(readRows, rasterBand->GetYSize() - i * readRows); 

rasterBand->RasterIO(GDALRWFlag::GF_Read, j * readCols, i * readRows, 
    nXSize, nYSize, rasterBlock, nXSize, nYSize, GDALDataType::GDT_CFloat32, 0, 0);

谢谢您的帮助。下面的代码正在工作，但是这样做比读取整个图像需要更多时间。我改变了 \t float * rasterBlock =（float *）CPLMalloc（nCols * nRows * sizeof（GDT_CFloat32））; \t float * rasterBlockOut =（float *）CPLMalloc（nCols * nRows * sizeof（GDT_CFloat32））; –

使用块的目的不是加速处理（如果处理整个图像），而是要处理大量不适合内存的数据。如果你可以在RAM中一次加载整个图像，它肯定会更快。 – Gwen

GDAL C++ RasterIO的块

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