CODEWITHSUNDEEP
cpointers
bs7280
bs7280

Reputation: 1094

Passing an array of pointers to strings into a function

I am trying to pass an array of pointers to strings (names) into a function (foo) and read from it. The below code produces a segmentation fault. Can someone please help me figure out why this code is resulting in a segmentation fault? I want to be able to pass the array names[][] through a function and work with the data like I would if I were using names[][] outside the function.

void foo(char *bar[]) {
    printf("%s\n", bar[0]);
}

//---------------Main-------------

char args[][50] = {"quick", "brown", "10", "brown", "jumps", "5"};
int i = 0;
int numbOfPoints = (sizeof(args)/sizeof(args[0]))/3;

//array of all the locations. the number will be its ID (the number spot in the array)
//the contents will be
char names[numbOfPoints][100];

for(i = 0; i < numbOfPoints; i++) {
    char *leadNode = args[i*3];
    char *endNode = args[i*3 + 1];
    char *length = args[i*3 + 2];
    int a = stringToInt(length);

    //add name
    strcpy(names[i],leadNode);
}

//printing all the names out
for(i = 0; i < numbOfPoints; i++) {
    printf("%s\n", names[i]);
}

foo(names);

Upvotes: 0

Views: 64

Answers (1)

R Sahu
R Sahu

Reputation: 206567

The problem is the the argument type of foo and the way you are calling it. The argument type of foo, char* [] is not compatible with name. I get the following warning in gcc 4.8.2 with -Wall.

soc.c:35:4: warning: passing argument 1 of ‘foo’ from incompatible pointer type [enabled by default]
    foo(names);
    ^
soc.c:5:6: note: expected ‘char **’ but argument is of type ‘char (*)[100]’
 void foo(char *bar[]) {

Change foo to:

void foo(char (*bar)[100]) {
    printf("%s\n", bar[0]);
}

and all should be well.

Upvotes: 1

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