Do not leak body in intermediate buffers to converters #3035
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Multiple calls to source() would create a new BufferedSource which, when a single byte was read, would cache data in its buffer which would be lost on a subsequent call to source(). Calls to source() will now always return the same BufferedSource ensuring data cannot be lost.
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Multiple calls to source() would create a new BufferedSource which, when a single byte was read, would cache data in its buffer which would be lost on a subsequent call to source(). Calls to source() will now always return the same BufferedSource ensuring data cannot be lost.
Closes #3034