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An efficient solution to convert a signed number character string to a signed int. Was going to put it up on Stack Overflow, but the question was closed as duplicate.
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| #include <stdio.h> | |
| #include <string.h> | |
| #include <math.h> | |
| int my_atoi(const char* snum) | |
| { | |
| int idx, strIdx = 0, accum = 0, numIsNeg = 0; | |
| const unsigned int NUMLEN = (int)strlen(snum); | |
| /* Check if negative number and flag it. */ | |
| if(snum[0] == 0x2d) | |
| numIsNeg = 1; | |
| for(idx = NUMLEN - 1; idx >= 0; idx--) | |
| { | |
| /* Only process numbers from 0 through 9. */ | |
| if(snum[strIdx] >= 0x30 && snum[strIdx] <= 0x39) | |
| accum += (snum[strIdx] - 0x30) * pow(10, idx); | |
| strIdx++; | |
| } | |
| /* Check flag to see if originally passed -ve number and convert result if so. */ | |
| if(!numIsNeg) | |
| return accum; | |
| else | |
| return accum * -1; | |
| } | |
| int main() | |
| { | |
| /* Tests... */ | |
| printf("Returned number is: %d\n", my_atoi("34574")); | |
| printf("Returned number is: %d\n", my_atoi("-23")); | |
| return 0; | |
| } |
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