[SOLVED] ESP8266 crash at if (HTTPCode > 0) {}

Projects Programming
1

Hello fellow programmers!

I am having trouble with my ESP, it seems to be crashing at an odd place.

I am using this function (I cut out some of the code, a lot of it is not relevant):

#include <ESP8266HTTPClient.h> // At the top of the project

String CollectExternalResponse(String URL) {
  WiFiClient client;
    
  HTTPClient HTTP;

  HTTP.begin(URL);
  int HTTPCode = HTTP.GET();

  Serial.println("A");
  String Response = "";
  Serial.println("B");

  if (HTTPCode > 0) {
    Serial.println("1");
    Response = HTTP.getString();
    Serial.println("2");
  }

  HTTP.end();
  return Response;
}

When I run it, the "A" and "B" print to the Serial monitor but "1" and "2" do not. The full output is this:

A
B

Exception (3):
epc1=0x40100621 epc2=0x00000000 epc3=0x00000000 excvaddr=0x40068648 depc=0x00000000

>>>stack>>>

ctx: cont
sp: 3ffffd80 end: 3fffffc0 offset: 01a0
3fffff20:  4010000a 002a8400 3ffee75c 3ffee700  
3fffff30:  3fffdad0 00000000 00000020 401008cb  
3fffff40:  00001388 3ffee75c 00000000 00000000  
3fffff50:  3ffef800 0035003f 3fffff80 4020424a  
3fffff60:  3fffff8c 3fffff8c 3ffee6c0 40204260  
3fffff70:  3fffdad0 00000000 3ffee6c0 402012ad  
3fffff80:  3ffe8648 8000001c 00000000 3ffef76c  
3fffff90:  0035003f 00000000 3ffee6c0 402010f6  
3fffffa0:  feefeffe feefeffe feefeffe 402051a8  
3fffffb0:  feefeffe feefeffe 3ffe84e8 40100b8d  
<<<stack<<<

 ets Jan  8 2013,rst cause:1, boot mode:(3,0)

load 0x4010f000, len 1392, room 16 
tail 0
chksum 0xd0
csum 0xd0
v3d128e5c
~ld

Anyone know what is going on here? It seems odd that it would crash at this if statement, especially given the same function works when I do not pass the URL as a variable. Just to be clear, in another test, I printed the URL at the beginning of the function and it worked.

Here is the URL: https://blockchain.info/tobtc?currency=USD&value=1000

I would really appreciate some help!

2

"Anyone know what is going on here? It seems odd that it would crash at this if statement, especially given the same function works when I do not pass the URL as a variable. Just to be clear, in another test, I printed the URL at the beginning of the function and it worked. "

If you hard code the URL in in place of the URL variable, does it work? What format/type is the URL variable?

3

Exception 3 is a stack overflow. Most likely the call to HTTP.getString() is returning a gigantic string, which causing the error.

Your print statements do NOT necessarily indicate WHERE the problem is, because it takes a long time for the characters to actually be sent out the serial port. If you follow each Serial.print with a Serial.flush(), the characters will all be sent before the next line of code is executed. If you do that, I expect you will see the "1" come out, but not the "2", which will confirm what I wrote above.

Regards,
Ray L.

4

Response = HTTP.getString();As Ray says, if the code isn't '0' then you should not attempt to do that.

5

RayLivingston:
Exception 3 is a stack overflow. Most likely the call to HTTP.getString() is returning a gigantic string, which causing the error.

Your print statements do NOT necessarily indicate WHERE the problem is, because it takes a long time for the characters to actually be sent out the serial port. If you follow each Serial.print with a Serial.flush(), the characters will all be sent before the next line of code is executed. If you do that, I expect you will see the "1" come out, but not the "2", which will confirm what I wrote above.

Regards,
Ray L.

Good suggestion, but that your prediction did not quite match the results. When I added Serial.flush() after each print, a bunch of gibberish was printed as well. The first character is discernible, but anything else is hidden in the spam.

If you follow the link I posted in the original thread, it leads to the conversion rate for Bitcoin in USD. It should be returning something like 0.14789, not a gigantic string (unless it is sending things in a different format... perhaps json?).

6

zoomkat:
"Anyone know what is going on here? It seems odd that it would crash at this if statement, especially given the same function works when I do not pass the URL as a variable. Just to be clear, in another test, I printed the URL at the beginning of the function and it worked. "

If you hard code the URL in in place of the URL variable, does it work? What format/type is the URL variable?

It does not. I am sending the variable to the function as a String type. When I define it as const char*, as shown in the ESP examples and is functional on a server I made myself to return predictable strings, I receive the same error.

7

UPDATE:

I changed the URL to "http://worldtimeapi.org/api/ip/75.110.25.187.txt" and had a successful response from the server.

There are two observations I have about this:

  1. It uses http instead of https. I don't know if this makes a difference, but I will test it to see.

  2. The URL ends in .txt. Though the original claims to be in plain text, this may be significant.

8

UPDATE:

Since the api uses HTTPS, I needed to specify a fingerprint in the code:

HTTP.begin(URL, Fingerprint);

This did not solve the stack overflow problem, though.

9

Have you tried to download the URL with a browser and store it to a file and see how big that file is?

10

Deva_Rishi:

Response = HTTP.getString();

As Ray says, if the code isn't '0' then you should not attempt to do that.

All of the examples I've seen specify the code should be greater than 0. A positive response from http is 200...

11

Danois90:
Have you tried to download the URL with a browser and store it to a file and see how big that file is?

I just tried something similar. I made the same request with python and used a function to see that the response was 48 bytes.

Is this similar enough to what you suggested?

12

"1) It uses http instead of https. I don't know if this makes a difference, but I will test it to see."

As best as I know, there is a lot of extra stuff you have to do with https.

13

The HTTPClient has no valid WiFi client applied - look at this example.

14

SOLVED

Ok! The issue was with this particular API and its need for https. The solution should apply to any plaintext API using https:

You need to specify the SHA-1 fingerprint in the code:

http.begin(URL, SHA1Fingerprint);

It is important to use the SHA-1 fingerprint! When I first tried this, I used SHA-256 which did not solve the problem.

How do you get the SHA-1 fingerprint, you may wonder! Here is what to do on a Mac (specifically Safari):

  1. In Safari, click on the green lock next to the url.

  2. Click "Show Certificate"

  3. Click "Details"

  4. At the very bottom you will find your SHA-1 fingerprint

My only question at this point is whether the fingerprint will eventually need to be updated. Will it ever expire? Will this happen in a predictable time frame?

My code is dynamic, so I can specify the fingerprint from the outside, but will I need to check in every so often to update that data?

15

The fingerprint will expire with the certificate, expiration date is shown in the certificate. If you properly implement SSL/TLS in you application, you do not need to worry about this.