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Sensacion Y Percepcion Schiffman Pdf Free [2022]







iphoneQ: Reading char in string I have a char array like this: char* a; char* b = "w"; char* c = "w"; a = b; a[0] = c[0]; c[0] = 'h'; After this I have printed a and b the result is: w wh I need to print the result like this: How can I do this? A: There's two problems with your code: This is not a valid definition of a pointer to a char. It should be char * a; The next line is invalid as well. You cannot assign a pointer to a char in C. You can assign it to a character. int main() { char * a = "w"; char * b = "w"; char * c = "w"; a[0] = c[0]; c[0] = 'h'; printf("a = %s, b = %s, c = %s ", a, b, c); return 0; } You could make this work if you wanted to, but I would not recommend it. And to answer your question, you need to print a, then b, then c. There's no magic trick here. char* a = "w"; printf("a = %s, b = %s, c = %s The printf() function expects a pointer to a char (what you gave it). Then it tries to printf() the char, which isn't valid. printf("%s ", a); This solution is better, because now you don't have to use array syntax, but strcpy() function. char a[100]; char b[100]; char c[100]; strcpy(a, "w"); strcpy


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