Convert const char* to binary operator+

Question

I'm stuck at trying to figure out the correct data types for T1 and P1 (see below).

Compiler error:

weather_station_v1_2.ino:76:37: error: invalid operands of types ‘const char*’ and ‘const char [11]’ to binary ‘operator+’

The Code:

if (status != 0) {
      delay(status);
      status = pressure.getPressure(P,T);

      const unsigned char P1 = pressure.getTemperature(P);
      const unsigned char T1 = pressure.getTemperature(T);

      byte data = "temp" + T1 + "&pressure=" + P1;
      httppost();
      delay(1000);
    }

At the end data will be send to a PHP server, but it has to follow the scheme name1=value1&name2=value2.

Would you be so nice and give me a hint about what I'm doing wrong? Thanks

are you in need of snprintf()?

Sourav Ghosh
– Sourav Ghosh

2016-06-22 18:14:25 +00:00
Commented Jun 22, 2016 at 18:14 — Sourav Ghosh
– Sourav Ghosh, Commented Jun 22, 2016 at 18:14
yes, it seems so xD

hehxes
– hehxes

2016-06-22 18:40:53 +00:00
Commented Jun 22, 2016 at 18:40 — hehxes
– hehxes, Commented Jun 22, 2016 at 18:40

datenwolf · Accepted Answer · 2016-06-22 19:42:13Z

2

Your problem is not one of data type conversion, but that in C the operator + is defined only for arithmetic operations (adding numbers, or adding an offset to a pointer).

You're attempting to concatenate strings. There is no "string" type built into C. C "strings" are char arrays, usually terminated with a 0 byte. Arrays in C are fixed size and fixed allocation so there is no array "concatenation" operator in C.

Operations on arrays, and thereby also on C "strings", are usually implemented through manipulation functions. In your particular case the usual approach is the use of a string formatting function of the printf family. You should look at snprintf.

edited Jun 22, 2016 at 19:42

answered Jun 22, 2016 at 18:20

datenwolf

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9 Comments

chux Over a year ago

Nice "hint" answer per OP request. UV (Quibble: Disagree with ""strings ... usually terminated with a 0 byte". In C, strings always end with a null character, else it is not a string.)

hehxes Over a year ago

Thanks for your help, my code does work now :)

if (status != 0) {           delay(status);           status = pressure.getPressure(P,T);                      const unsigned char P1 = pressure.getTemperature(P);           const unsigned char T1 = pressure.getTemperature(T);            char buffer [100];           byte data = snprintf(buffer, 100, "temp=%T1&pressure=%P1", T1, P1);                      httppost();           delay(1000);         }

chux Over a year ago

@hehxes "temp=%T1&pressure=%P1" is an unusual format. Sure it works?

datenwolf Over a year ago

@chux: Yes, the C standard defines that const string literals that are not initializers always terminate with a 0 byte. However this char hello[5] = "hello"; is perfectly legal, but will not terminate with a 0 byte. As far as the standard library str… functions is concerned this is not a string of course. The borders of what a string is and what not are a bit blurry. IMHO the use of the in-band signalling 0 byte terminator was one of the worse decisions in the C design process. Sure, having to drag around the length explicitly is annoying, but it would have prevented so many security bugs.

chux Over a year ago

"hello" is a string literal, not necessarily a string. True C has blurry terminology.

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