尝试增加字符的ascii值时，为什么会得到随机数？

Question

int n = argv[i][j];
n = n + (int)argv[1]; 

/* I'm pretty sure the above part that is wrong. What I was hoping that this 
would do is take the number the person put in and increase n by that number but 
it isnt working*/ 

printf("%d\n", n);

• i = the string numbers of the argument
• j = the characters of the string

What I want is to make it so when someone types ./123 12 hi I want it to increase the ascii characters of h and i by 12 or whatever number they put in. When I test my code out with

./123 1 hi 

I get an output of

-1081510229
-1081510228

instead of

105
106 

which is i and j , the next letters of h and i

Libraries I'm using

stdio.h 
studio50.h
string.h

Answer 1

argv[1] is a string (ie, a null-terminated char array), casting it to int doesn't give the result you expected

You should use atoi to do the job. Or better, use strtol to get better stability.