version 0.5.1, no exception but break the string line when dump some special data

when I use this lib to dumps some data, it make a puzzled question, [sample.log](https://github.com/hhatto/pyrapidjson/files/1634053/sample.log) is the minimized data to show it. my python version is 2.7.13 .

1. pyrapidjson 0.5 + not encode with utf-8
```
# -*- coding:utf-8 -*-
import codecs
import rapidjson
if __name__ == '__main__':
with codecs.open('sample.log', 'rb', 'utf-8') as f_read:
report_data = [f_read.read(),]
print report_data
report_json = rapidjson.dumps(report_data)
print report_json
```
it raise a crash about python like:


2. pyrapidjson 0.5 + encode with utf-8
```
# -*- coding:utf-8 -*-
import codecs
import rapidjson
if __name__ == '__main__':
with codecs.open('sample.log', 'rb', 'utf-8') as f_read:
report_data = [f_read.read().encode('utf-8'),]
print report_data
report_json = rapidjson.dumps(report_data)
print report_json
```
it may right like:


3. pyrapidjson 0.5.1 + not encode with utf-8
```
# -*- coding:utf-8 -*-
import codecs
import rapidjson
if __name__ == '__main__':
with codecs.open('sample.log', 'rb', 'utf-8') as f_read:
report_data = [f_read.read(),]
print report_data
report_json = rapidjson.dumps(report_data)
print report_json
```
it like the following picture, no exception but break the string line:


4. pyrapidjson 0.5.1 + encode with utf-8
```
# -*- coding:utf-8 -*-
import codecs
import rapidjson
if __name__ == '__main__':
with codecs.open('sample.log', 'rb', 'utf-8') as f_read:
report_data = [f_read.read().encode('utf-8'),]
print report_data
report_json = rapidjson.dumps(report_data)
print report_json
```
it like the following picture, also no exception but break the string line:


with this condition, I can't know right or not about dumps result. I think it may raise an exception when meet this condition or another way to let user know about this error.