Passing char* to atoi

Question

Here's the scenrio -

        #include <stdio.h>
        #include <stdlib.h>
        #include <stdint.h>

        int main()
        {
            uint8_t backoff;
            char* value = "300000";
            backoff=atoi(value);
            printf("value = %s\n", value);
            printf("backoff value = %d\n", backoff);

            return (0);
        }

Output is as -

        value = 300000
        backoff value = 224

Can someone please help me understand how this conversion happened ?

300000 won't fit in `uint8_t`. – Eugene Sh. Jun 09 '21 at 16:53 — Eugene Sh., Jun 09 '21 at 16:53
300000 == 0x493e0. 224 == 0xe0 – stark Jun 09 '21 at 16:56 — stark, Jun 09 '21 at 16:56

score 0 · Accepted Answer · answered Jun 09 '21 at 16:55

0

Pretty simple. Values of uint8_t are 0 to 255. You need at least uint32_t (uint16_t max 65535). 224 is the bits of the real answer that fit in the 8 bit int.

answered Jun 09 '21 at 16:55

zBeeble

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Passing char* to atoi

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