problem with argv in c

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problem with argv in c

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#1 by (2 votes)

1

I'm writing a program to go through one of the arguments and store them in a variable, but I get a segmentation error right in this loop:

 for(int i=0; argv[1][i] != " for(int i=0; argv[1][i] != "%pre%" ;i++){
    }
" ;i++){
    }

How else could argv walk to avoid this error?

c

asked by kalia 02.05.2017 в 19:25

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1 answer

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score 2 · Accepted Answer

You make an incorrect comparison. Surely, the compiler will give you a warning ( warning ).

argv[1][i] != "char" are comparing a argv[1][i] with a pointer . Well, actually the compiler is promoting int to int , so, actually, you're comparing a "..." with a pointer .

For what you want to do, you can use

argv[1][i] != 0

or

argv[1][i] != 'argv[1][i] != 0
' // Comillas simples '

Remember that, in C, strings enclosed in double quotes %code% are no more than arrays of characters. By using them, you actually use your address , that is, a pointer .