You can use the following function to replace the text of one buffer with the text of another buffer:
This function replaces the accessible portion of the current buffer
with the accessible portion of the buffer source. source
may either be a buffer object or the name of a buffer. When
replace-buffer-contents succeeds, the text of the accessible
portion of the current buffer will be equal to the text of the
accessible portion of the source buffer.
This function attempts to keep point, markers, text properties, and
overlays in the current buffer intact. One potential case where this
behavior is useful is external code formatting programs: they
typically write the reformatted text into a temporary buffer or file,
would destroy these properties. However, the latter combination is
typically faster (See Deleting Text, and Inserting Text).
For its working,
replace-buffer-contents needs to compare the
contents of the original buffer with that of source which is a
costly operation if the buffers are huge and there is a high number of
differences between them. In order to keep
replace-buffer-contents’s runtime in bounds, it has two
max-secs defines a hard boundary in terms of seconds. If given
and exceeded, it will fall back to
max-costs defines the quality of the difference computation. If the actual costs exceed this limit, heuristics are used to provide a faster but suboptimal solution. The default value is 1000000.
t if a non-destructive
replacement could be performed. Otherwise, i.e., if max-secs
was exceeded, it returns
This function replaces the region between beg and end using the given replace-fn. The function replace-fn is run in the current buffer narrowed to the specified region and it should return either a string or a buffer replacing the region.
The replacement is performed using
above) which also describes the max-secs and max-costs
arguments and the return value.
Note: If the replacement is a string, it will be placed in a temporary
buffer so that
replace-buffer-contents can operate on it.
Therefore, if you already have the replacement in a buffer, it makes
no sense to convert it to a string using