I want to save the processed image from OpenCV as the `input_name+enhanced.input_image_format`

Question

Let assume my input image name is cat.jpg after I processed the image I want to save the output image as cat_enhanced.jpg. How do I can do like that?

I don't know how to access the image name and format separately.

char* inputImgName = argv[1];

This code store both image name and format in the inputImgName pointer.

[Edited]

std::string imgNameString = argv[1];

Using this I can get the cat.jpg, but don't know how to save the image as described above.

Expected output:
input image name: cat.jpg
output image name: cat_enhanced.jpg

You'd be better off with a std::string to handle the memory allocation and reallocation for you. But if you want to use a char pointer, you'll have to allocate the memory yourself; you shouldn't point it to argv[1] and munge that memory. — L. Scott Johnson, Jun 13 '19 at 13:49
@L.ScottJohnson I edited my question. Do you know how to save the image as I asked above. — hua, Jun 13 '19 at 13:57
See https://stackoverflow.com/questions/5878775/how-to-find-and-replace-string — L. Scott Johnson, Jun 13 '19 at 13:57

score 0 · Accepted Answer · answered Jun 13 '19 at 14:18

This is my answer to my question. Thanks @ L. Scott Johnson for pointing me out.

std::string inputImgName = argv[1];
std::string findFormat(".");
size_t pos = inputImgName.find(findFormat);
inputImgName.replace(pos, findFormat.length(), "_enhanced.");
cout << "The enhanced image name is: " << inputImgName << endl;

I want to save the processed image from OpenCV as the `input_name+enhanced.input_image_format`

1 Answers1