Covert char[2] to uint16_t for socket programming?

Question

I am writing a tftp client. But when i convert the block number as follows:

uint16_t blockN = buffer[2]<<8 | buffer[3];

after 127, i am getting 65408 as blockN. What might be the problem here?

Thank you for your answers.

i don't remember operator precedence table, but try to use brackets, like `(buffer[2]<<8) | (buffer[3]);` — Iłya Bursov, Nov 29 '14 at 00:26

score 4 · Answer 1 · answered Nov 29 '14 at 00:25

4

You have to change the type of buffer array from an array of char to an array of unsigned char, otherwise buffer[2] will be promoted to int and sign extension will occur. On most platforms char type is a signed type.

answered Nov 29 '14 at 00:25

ouah

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Ok i solved but not exaclty this way but this idea helped. I did uint16_t blockN = buffer[2]<<8 | buffer[3]&0xFF; so i got rid of sign extensions – ahmetsarias Nov 29 '14 at 00:36

score 0 · Accepted Answer · answered Nov 29 '14 at 00:39

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I solved by doing uint16_t blockN = buffer[2]<<8 | (buffer[3]&0xFF);

answered Nov 29 '14 at 00:39

ahmetsarias

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Covert char[2] to uint16_t for socket programming?

2 Answers2